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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Interesting inequality
sqing   0
a few seconds ago
Source: Own
Let $ a,b\geq 2  . $ Prove that
$$(1-a^2)(1-b^2) -6ab\geq-15$$$$(1-a^2)(1-b^2) -7ab\geq  -\frac{58}{3}$$$$(1-a^3)(1-b^3) -\frac{21}{4}a^2b^2\geq -35$$$$(1-a^3)(1-b^3) -7a^2b^2\geq-\frac{2391}{49}$$
0 replies
2 viewing
sqing
a few seconds ago
0 replies
Find max no. of gangsters
sk2005   4
N 2 minutes ago by flower417477
In Chicago, there are 36 criminal gangs, some of which are at war with
each other. Each gangster belongs to several gangs and every pair of gangsters
belongs to a different set of gangs. It is known that no gangster is a member of
two gangs which are at war with each other. Furthermore, each gang that some
gangster does not belong to is at war with some gang he does belong to. What is
the largest possible number of gangsters in Chicago?
4 replies
sk2005
Sep 13, 2021
flower417477
2 minutes ago
super duper ez radax problem
iStud   1
N 3 minutes ago by MathLuis
Source: Monthly Contest KTOM March 2025 P1 Essay
Given an acute triangle $ABC$ with $BC<AB<AC$. Points $D$ and $E$ are on $AB$ and $AC$ respectively such that $DB=BC=CE$. Lines $CD$ and $BE$ meet at $F$. $I$ is the incenter of $\triangle{ABC}$ and $H$ is the orthocenter of $\triangle{DEF}$. $\omega_b$ and $\omega_c$ are circles with diameter $BD$ and $CE$, respectively, intersecting each other at points $X$ and $Y$. Prove that $I$ and $H$ lie on $XY$.

Hint
1 reply
1 viewing
iStud
27 minutes ago
MathLuis
3 minutes ago
how do we find a construction?
iStud   0
18 minutes ago
Source: Monthly Contest KTOM March 2025 P4 Essay
Given a chess board $n\times n$ with $n>3$ with all the unit squares are initially white coloured. Every move, we can turn the color (from white to black or otherwise) from the 5 unit squares that form this T-pentomino which can be rotated or reflexed (see the image below). Determine all natural numbers $n$ such that all unit squares on the board can be made into all black after a finite number of moves.
0 replies
iStud
18 minutes ago
0 replies
unnecessary wrapped FE on Q
iStud   0
22 minutes ago
Source: Monthly Contest KTOM March 2025 P3 Essay
Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that
\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]for all rational numbers $x,y$.

Hint
0 replies
+1 w
iStud
22 minutes ago
0 replies
surprisingly not trivial for a P2
iStud   0
25 minutes ago
Source: Monthly Contest KTOM March 2025 P2 Essay
Find all natural numbers $(m,n)$ such that
\[2^{n!}+1\mid 2^{m!}+19\]
Hint
0 replies
iStud
25 minutes ago
0 replies
Inspired by Mihaela Berindeanu
sqing   2
N 30 minutes ago by sqing
Source: Own
Let $ a,b,c>0  . $ Prove that
$$(4a+b+c)(a+b)(b+c)(c+a)\geq (ab+bc+ca)(2a+b+c)^2$$
2 replies
sqing
Yesterday at 2:08 PM
sqing
30 minutes ago
The Sums of Elements in Subsets
bobaboby1   1
N 35 minutes ago by bobaboby1
Given a finite set \( X = \{x_1, x_2, \ldots, x_n\} \), and the pairwise comparison of the sums of elements of all its subsets (with the empty set defined as having a sum of 0), which amounts to \( \binom{2}{2^n} \) inequalities, these given comparisons satisfy the following three constraints:

1. The sum of elements of any non-empty subset is greater than 0.
2. For any two subsets, removing or adding the same elements does not change their comparison of the sums of elements.
3. For any two disjoint subsets \( A \) and \( B \), if the sums of elements of \( A \) and \( B \) are greater than those of subsets \( C \) and \( D \) respectively, then the sum of elements of the union \( A \cup B \) is greater than that of \( C \cup D \).

The question is: Does there necessarily exist a positive solution \( (x_1, x_2, \ldots, x_n) \) that satisfies all these conditions?
1 reply
bobaboby1
Mar 12, 2025
bobaboby1
35 minutes ago
Max amount of equal numbers among (a_i^2 + a_j^2)/(a_i + a_j)
mshtand1   1
N an hour ago by Rushery_10
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 9.8
Given $2025$ pairwise distinct positive integer numbers \(a_1, a_2, \ldots, a_{2025}\), find the maximum possible number of equal numbers among the fractions of the form
\[
\frac{a_i^2 + a_j^2}{a_i + a_j}
\]
Proposed by Mykhailo Shtandenko
1 reply
mshtand1
Mar 14, 2025
Rushery_10
an hour ago
Combinatorics from Iranian TST 2017
bgn   20
N an hour ago by ezpotd
Source: Iranian TST 2017, first exam, day1, problem 2
In the country of Sugarland, there are $13$ students in the IMO team selection camp.
$6$ team selection tests were taken and the results have came out. Assume that no students have the same score on the same test.To select the IMO team, the national committee of math Olympiad have decided to choose a permutation of these $6$ tests and starting from the first test, the person with the highest score between the remaining students will become a member of the team.The committee is having a session to choose the permutation.
Is it possible that all $13$ students have a chance of being a team member?

Proposed by Morteza Saghafian
20 replies
bgn
Apr 5, 2017
ezpotd
an hour ago
C-B=60 <degrees>
Sasha   26
N an hour ago by shendrew7
Source: Moldova TST 2005, IMO Shortlist 2004 geometry problem 3
Let $O$ be the circumcenter of an acute-angled triangle $ABC$ with ${\angle B<\angle C}$. The line $AO$ meets the side $BC$ at $D$. The circumcenters of the triangles $ABD$ and $ACD$ are $E$ and $F$, respectively. Extend the sides $BA$ and $CA$ beyond $A$, and choose on the respective extensions points $G$ and $H$ such that ${AG=AC}$ and ${AH=AB}$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if ${\angle ACB-\angle ABC=60^{\circ }}$.

Proposed by Hojoo Lee, Korea
26 replies
Sasha
Apr 10, 2005
shendrew7
an hour ago
this hAOpefully shoudn't BE weird
popop614   46
N 2 hours ago by hgomamogh
Source: 2023 IMO Shortlist G1
Let $ABCDE$ be a convex pentagon such that $\angle ABC = \angle AED = 90^\circ$. Suppose that the midpoint of $CD$ is the circumcenter of triangle $ABE$. Let $O$ be the circumcenter of triangle $ACD$.

Prove that line $AO$ passes through the midpoint of segment $BE$.
46 replies
popop614
Jul 17, 2024
hgomamogh
2 hours ago
postaffteff
JetFire008   9
N 2 hours ago by drago.7437
Source: Internet
Let $P$ be the Fermat point of a $\triangle ABC$. Prove that the Euler line of the triangles $PAB$, $PBC$, $PCA$ are concurrent and the point of concurrence is $G$, the centroid of $\triangle ABC$.
9 replies
JetFire008
Mar 15, 2025
drago.7437
2 hours ago
A scary fish and a fiend
nukelauncher   96
N 2 hours ago by Mathandski
Source: USA November TST for IMO 2021 and TST for EGMO 2021, Problem 2, by Zack Chroman and Daniel Liu
Let $ABC$ be a scalene triangle with incenter $I$. The incircle of $ABC$ touches $\overline{BC},\overline{CA},\overline{AB}$ at points $D,E,F$, respectively. Let $P$ be the foot of the altitude from $D$ to $\overline{EF}$, and let $M$ be the midpoint of $\overline{BC}$. The rays $AP$ and $IP$ intersect the circumcircle of triangle $ABC$ again at points $G$ and $Q$, respectively. Show that the incenter of triangle $GQM$ coincides with $D$.

Zack Chroman and Daniel Liu
96 replies
nukelauncher
Nov 16, 2020
Mathandski
2 hours ago
Functional Inequality Implies Uniform Sign
peace09   30
N Yesterday at 7:09 AM by Nari_Tom
Source: 2023 ISL A2
Let $\mathbb{R}$ be the set of real numbers. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that \[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]for every $x,y\in\mathbb{R}$. Assume that the inequality is strict for some $x_0,y_0\in\mathbb{R}$.

Prove that either $f(x)\geqslant 0$ for every $x\in\mathbb{R}$ or $f(x)\leqslant 0$ for every $x\in\mathbb{R}$.
30 replies
peace09
Jul 17, 2024
Nari_Tom
Yesterday at 7:09 AM
Functional Inequality Implies Uniform Sign
G H J
G H BBookmark kLocked kLocked NReply
Source: 2023 ISL A2
The post below has been deleted. Click to close.
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peace09
5411 posts
#1 • 4 Y
Y by OronSH, MarkBcc168, Rounak_iitr, Sedro
Let $\mathbb{R}$ be the set of real numbers. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that \[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]for every $x,y\in\mathbb{R}$. Assume that the inequality is strict for some $x_0,y_0\in\mathbb{R}$.

Prove that either $f(x)\geqslant 0$ for every $x\in\mathbb{R}$ or $f(x)\leqslant 0$ for every $x\in\mathbb{R}$.
This post has been edited 2 times. Last edited by peace09, Jul 17, 2024, 12:27 PM
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peace09
5411 posts
#2 • 3 Y
Y by OronSH, Sedro, TensorGuy666
First, letting $y=x$ gives $f(0)f(2x)\ge0$, which implies the result unless $f(0)=0$; henceforth suppose so.

Next, letting $y=-x$ produces $0\ge f(x)^2-f(-x)^2$ or $f(x)^2\le f(-x)^2$; but swapping $x\mapsto-x$ analogously yields $f(-x)^2\le f(x)^2$, and so $f(x)^2=f(-x)^2~(\ast)$.

Then, letting $(x,y)=(x,y),(y,x)$ gives
\begin{align*}
        f(x+y)f(x-y)&\ge f(x)^2-f(y)^2\\
        f(y+x)f(y-x)&\ge f(y)^2-f(x)^2,
    \end{align*}and chaining the $1^\text{st}$ inequality with the negation of the $2^\text{nd}$ produces
\begin{align*}
        f(x+y)f(x-y)\ge-f(x+y)f(y-x)\\
        \iff f(w)f(z)\ge-f(w)f(-z),~(\dagger)
    \end{align*}where in the last step we only redenote $(x+y,x-y)\mapsto(w,z)$ for simplicity.

Now, it is given that $(\dagger)$ is strict for some $(\tilde{w},\tilde{z})$: \[f(\tilde{w})f(\tilde{z})>-f(\tilde{w})f(-\tilde{z}).\]But $(\ast)$ implies that the magnitudes of both sides are equal, and so both $f(\tilde{w})f(\tilde{z})$ and $f(\tilde{w})f(-\tilde{z})$ are strictly positive. In particular, $f(\tilde{z})=f(-\tilde{z})\neq0$.

Finally, letting $z=\tilde{z}$ in $(\dagger)$ yields $f(w)f(\tilde{z})\ge-f(w)f(-\tilde{z})=-f(w)f(\tilde{z})$, i.e., each $f(w)$ has the same sign as $f(\tilde{z})$. $\square$
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OronSH
1720 posts
#3 • 1 Y
Y by peace09
Call the assertion $P(x,y).$ If $P(a,b)$ is strict, then $P(a,b)+P(b,a)$ gives $f(a+b)(f(a-b)+f(b-a))>0,$ so $f(a-b)+f(b-a)\ne 0.$

Now $P\left(\frac{x+a-b}2,\frac{x-a+b}2\right)+P\left(\frac{x-a+b}2,\frac{x+a-b}2\right)$ gives $f(x)(f(a-b)+f(b-a))\ge 0.$ If $f(a-b)+f(b-a)>0$ then $f(x)\ge 0$ for all $x,$and if $f(a-b)+f(b-a)<0$ then $f(x)\le 0$ for all $x.$
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Anzoteh
124 posts
#4 • 20 Y
Y by OronSH, peace09, ehuseyinyigit, kamatadu, navi_09220114, pingupignu, Seicchi28, avisioner, Supertinito, BlazingMuddy, Sedro, CaptainLevi16, SatisfiedMagma, Aryan-23, EpicBird08, alexanderhamilton124, NicoN9, CyclicISLscelesTrapezoid, aidan0626, MS_asdfgzxcvb
I am the first proposer of this problem :) Co-authored with Ivan Chan (who proposed IMO 2023 P3 which imho a lot better than this), and Tristan Chaang (who was the first to supplya clean and neat solution; the original solution by Ivan and me made it looked like it's an A5).

I originally thought of the problem with sign flipped, i.e. $f(x + y) f(x - y) \le f(x)^2 - f(y)^2$, and then realized that leads to equality (i.e. no room on what to do). Then I decided to try the other direction and took me a while to notice the fact that became the problem statement!

Unfortunately I can't further characterize such functions. (What happens if we have $f(x + y) f(x - y) = f(x)^2 - f(y)^2$ and $f(x_0), f(y_0)$ different signs? How about the case where $f(x + y) f(x - y) > f(x)^2 - f(y)^2$ for all $x, y$? Those characterization seems elusive. )
This post has been edited 1 time. Last edited by Anzoteh, Jul 17, 2024, 12:12 PM
Reason: Typo
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kamatadu
466 posts
#5 • 1 Y
Y by SilverBlaze_SY
Thanks to Sammy27 for pointing out the point-wise trap in my previous (fake-)solution :ninja: . Here is the previous solution if anyone wants. incorrect solution

Here is the corrected solution.


Let $P(x,y)$ denote the assertion.

$P(x,x)\implies f(2x)f(0) \ge 0$.

Now if $f(0) \neq 0 $, then we can just divide both sides by $f(0)$ and switch $x \mapsto \frac{x}{2}$ to get that $f(x) \ge 0$ or $f(x) \le 0$ for all $x$.

Otherwise, assume $f(0) = 0$.

$P(x,-x) \implies 0 \ge f(x)^2 - f(-x)^2 \implies f(-x)^2 \ge f(x)^2$.

Now switching $x \mapsto -x$ in the above inequality, we get that $f(x)^2 \ge f(-x)^2$. Adding these two up, we get $f(-x)^2 + f(x)^2 \ge f(x)^2 + f(-x)^2$ which forces that equality holds in both the cases, i.e., $f(x)^2 = f(-x)^2$ for all $x$.

\[
P(x,y) \implies f(x+y)f(x-y) \ge f(x)^2 - f(y)^2.
\]\[
P(-y,x) \implies f(-y + x)f(-y-x) \ge f(-y)^2 - f(x)^2 = f(y)^2 - f(x)^2.
\]Adding these two up, we get that,
\[
f(x-y)(f(x+y) + f(-(x+y))) \ge 0
.\]
Now we substitute $x \mapsto \frac{p+q}{2}$ and $y \mapsto \frac{q-p}{2}$ to get,
\[
f(p)(f(q) + f(-q)) \ge 0
.\]
Now if $f(q)+f(-q) \neq 0$ for some $q$, then we are basically done. FTSOC assume that $f(q) = -f(-q)$ for all $q$, i.e., $f$ is odd.

Then,
\[
P(x,y) \implies f(x+y)f(x-y) \ge f(x)^2 - f(y)^2
.\]
\[
P(y,x) \implies f(y+x)f(y-x) \ge f(y)^2 - f(x)^2 \implies -f(x+y)f(x-y) \ge f(y)^2 - f(x)^2
.\]
Adding these two, we get that $0\ge 0$ which forces the equality of both the equations, that is
\[
f(x+y)f(x-y) = f(x)^2 - f(y)^2
\]for all $x$, $y$. But this is a contradiction because for $(x_0,y_0)$ the equality fails and we are done.



Can someone clarify what the $\gg$ means?

@below, thanks :thumbup:
This post has been edited 7 times. Last edited by kamatadu, Jul 18, 2024, 2:15 PM
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peace09
5411 posts
#6
Y by
kamatadu wrote:
Can someone clarify what the $\gg$ means? I have a solution that seems to be pretty weird :maybe: . Will be posting soon.
Sorry, I meant $\geqslant$, non-strict inequality.
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OptimalFEian
10 posts
#7
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Let $P(x, y)$ denote the given assertion. First, $P(x/2, x/2)$ gives $f(x)f(0) \geq 0$ for all $x$. Thus, we are done if $f(0) \neq 0$. Now, assume $f(0) = 0$. Then, $P(x, -x)$ and $P(-x, x)$ give
\[f(-x)^2 \geq f(x)^2 \geq f(-x)^2 \implies f(x)^2 = f(-x)^2 \enspace \forall x \in \mathbb{R}.\]We have
\begin{align*}
		P(\frac{x+y}{2}, \frac{x-y}{2}): f(x)f(y) \geq f\left(\frac{x+y}{2} \right)^2 - f\left(\frac{x-y}{2} \right)^2
	\end{align*}for all $x, y \in \mathbb{R}$, and denote this by $Q(x, y)$. Suppose $f(-a) = -f(a)$ for some $a$. Then $Q(x, a)$ and $Q(x, -a)$ yield
\begin{align*}
		f(x)f(-a) =f\left(\frac{x-a}{2} \right)^2 - f\left(\frac{x+a}{2} \right)^2 \enspace \forall x \in \mathbb{R}.
	\end{align*}Again, replacing $x$ by $-x$ gives $f(-x) f(-a) = -f(x)f(-a)$. It follows that $f(-x) = -f(x)$ for all real $x$, assuming that $f(a) \neq 0$. Then, for any $x, y \in \mathbb{R}$,
\begin{align*}
		f(y+x)f(y-x)\geq f(y)^2 - f(x)^2 \geq -f(x+y)f(x-y)=f(x+y)f(y-x)
	\end{align*}which forces $P(x, y)$ to be the equality, contradicting the hypothesis.
Hence, we now have $f(-x) = f(x)$ for all $x$. It follows that
\begin{align*}
		f(x)f(-y) \geq f\left(\frac{x-y}{2} \right)^2 - f\left(\frac{x+y}{2} \right)^2 \geq -f(x)f(y).
	\end{align*}That is, $f(x)f(y) \geq 0$ for all real $x, y$. Now, the conclusion follows.
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Marinchoo
407 posts
#8 • 1 Y
Y by isomoBela
Denote by $P(x,y)$ the assertion of $(x,y)$ into the functional equation. Combining the strict $P(x_0, y_0)$ and $P(y_0, x_0)$, we get:
\begin{align*}
    f(x_0+y_0)f(x_0-y_0)&>f(x_0)^2-f(y_0)^2 \\
    f(x_0+y_0)f(y_0-x_0)&\geq f(y_0)^2-f(x_0)^2 \\
    \Longrightarrow f(x_0+y_0)\cdot (f(x_0-y_0)+f(y_0-x_0)) &> 0.
\end{align*}Note that $f$ works iff $-f$ does, so WLOG $f(s)+f(-s)>0$ where $s = x_0-y_0$. Summing $P(x+s,x)$ and $P(x,x+s)$ now shows $f$ is non-negative, as desired.
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ihatemath123
3429 posts
#9 • 1 Y
Y by TensorGuy666
Let $(a,b)$ be a pair that makes the inequality strict. Plugging in the pairs $(a,b)$ and $(b,a)$ and summing gives us
\[ f(a+b) [ f(a-b) + f(b-a) ] > 0 \implies f(a-b)+f(b-a) \neq 0.\]Let $k$ be any constant; plugging in the pairs $(a+k, b+k)$ and $(b+k, a+k)$ and summing gives us
\[ f(a+b+2k) [ f(a-b) + f(b-a) ] \geq 0 \implies \text{sgn} (f(a+b+2k)) = \text{sgn} (f(a-b)+f(b-a))\]Since $k$ varies, $a+b+2k$ can be any real number, so we're done.
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sami1618
871 posts
#10
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We will prove the converse that $f$ achieving both signs implies that the inequality is an equality.

Claim: $f(a)$ and $f(-a)$ are of different signs (if they are not both zero)
Let $b$ be such that $f(a)$ and $f(b)$ have different signs. Then choose $x$ and $y$ such that $x+y=b$ and $x-y=a$. Then $P(x,y)$ gives that $0>f(x)^2-f(y)^2$ so $P(y,x)$ gives that $f(b)f(-a)$ must be positive, as desired.
Claim: $f(0)=0$
If $f(0)\neq 0$ then $P(x,x)$ would contradict our assumption.
Claim: $f(a)=-f(-a)$
Compare $P(a,-a)$ and $P(-a,a)$.

To finish comparing $P(x,y)$ and $P(y,x)$ shows that the inequality must be an equality.
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dkedu
179 posts
#11
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Let $P(x,y)$ denote the assertion.

Rewrite the condition as
\[-f(x+y)f(y-x) \le f(x)^2 - f(y)^2 \le f(x+y)f(x-y)\]\[-f(x+y)f(y-x) \le f(x+y)f(x-y)\]
Case 1: If there is $r$ such that $f(r) = f(-r) \neq 0$, then we have that by $P(\frac{n+r}{2},\frac{n-r}{2})$, which gives that $-f(n)f(-r) \le f(n)f(r)$ which implies that $f(n)$ has the same sign as $f(r)$ which implies $f(x) \le 0$ or $f(x) \ge 0$.

Case 2: If $f(x) = -f(-x)$ for all $x \in \mathbb R$, then we get that $f(x)^2 - f(y)^2 = f(x+y)f(x-y)$ so the inequality is never strict.

Having exhausted all cases, we are done.
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Sammy27
81 posts
#12 • 1 Y
Y by Eka01
Let $P(x, y)$ denote the assertion
$$f(x+y)f(x-y)\geq f(x)^2-f(y)^2.$$
  • $P(x_0,y_0)$ and $P(y_0, x_0)$ give
    $$f(x_0+y_0)f(x_0-y_0)> f(x_0)^2-f(y_0)^2$$and
    $$f(x_0+y_0)f(y_0-x_0)\geq f(y_0)^2-f(x_0)^2,$$respectively. We add them to get
    $$f(x_0+y_0)(f(x_0-y_0)+f(y_0-x_0))>0\implies f(x_0-y_0)+f(y_0-x_0)\neq 0.$$
  • $P\left(\frac{x+x_0-y_0}{2}, \frac{x+y_0-x_0}{2}\right)$ and $P\left(\frac{x+y_0-x_0}{2}, \frac{x+x_0-y_0}{2}\right)$ give
    $$f(x)f(x_0-y_0)\geq f\left(\frac{x+x_0-y_0}{2}\right)^2-f\left(\frac{x+y_0-x_0}{2}\right)^2$$and
    $$f(x)f(y_0-x_0)\geq f\left(\frac{x+y_0-x_0}{2}\right)^2-f\left(\frac{x+x_0-y_0}{2}\right)^2,$$respectively. We add them to get
    $$f(x)(f(x_0-y_0)+f(y_0-x_0))\geq 0,$$so
    $$f(x_0-y_0)+f(y_0-x_0)>0 \implies f(x)\geq 0$$or
    $$f(x_0-y_0)+f(y_0-x_0)<0 \implies f(x)\leq 0$$for all $x\in\mathbb{R}$ as desired, and we are done. $\blacksquare$
This post has been edited 2 times. Last edited by Sammy27, Jul 18, 2024, 5:01 PM
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crocodilepradita
145 posts
#13
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Let $P(x,y)$ be the assertion of $f(x+y)f(x-y)\ge f(x)^2-f(y)^2$.

$P(x_0,y_0) \implies f(x_0+y_0)f(x_0-y_0)>f(x_0)^2-f(y_0)^2$
$P(y_0,x_0) \implies f(x_0+y_0)f(y_0-x_0)\ge f(y_0)^2-f(x_0)^2$

Adding those $2$ inequality yields :
$f(x_0+y_0)[f(x_0-y_0)+f(y_0-x_0)]>0$ ...$(1)$

Claim 1. If $f(0)=0$ then $f(x)^2=f(-x)^2$ for all $x\in \mathbb{R}$
Proof.
Using $P(x,x)$ we get $f(2x)f(0)\ge 0$.
1) If $f(0)>0$ thus $f(2x)\ge 0 \implies f(x)\ge 0$ for all $x\in \mathbb{R}$.
2) If $f(0)<0 $ thus $f(2x)\le 0 \implies f(x) \le 0$ for all $x \in \mathbb{R}$.
3) If $f(0)=0$ then
$P(x,-x) \implies f(x)^2\le f(-x)^2$
$P(-x,x) \implies f(x)^2\ge f(-x)^2$
Therefore, $f(x)^2=f(-x)^2 ,\forall x \in \mathbb{R}$. $\square$

By substituting $x \to x_0-y_0$ we have
$f(x_0-y_0)^2=f(y_0-x_0)^2$
Therefore, $f(x_0-y_0)=f(y_0-x_0)$ or $f(x_0-y_0)=-f(y_0-x_0)$.

Now, consider inequality $(1)$.

If $f(x_0-y_0)=-f(y_0-x_0) \implies 0>0$, contradiction. Therefore, $f(x_0-y_0)=f(y_0-x_0)$.

Claim 2. If there exist $a\in \mathbb{R}$ such that $f(a)=f(-a)\neq 0$, then $f(x)\ge 0$ or $f(x)\le 0$ for all $x \in \mathbb{R}$
Proof.
$P(\frac{a+b}{2},\frac{a-b}{2}) \implies f(a)f(b)\ge f(\frac{a+b}{2})^2-f(\frac{a-b}{2})^2$
$P(\frac{-a+b}{2},\frac{-a-b}{2}) \implies f(-a)f(b) \ge f(\frac{b-a}{2})^2-f(\frac{-a-b}{2})^2=f(\frac{a-b}{2})^2-f(\frac{a+b}{2})^2$
Adding those $2$ inequality yields :
$f(b)[f(a)+f(-a)]\ge 0$

Now, if $f(a)=f(-a)>0$ then $f(b)\ge 0, \forall b \in \mathbb{R} \implies f(x)\ge 0, \forall x \in \mathbb{R} $. If $f(a)=f(-a)<0$ then $f(b)\le 0, \forall b \in \mathbb{R} \implies f(x)\le 0, \forall x \in \mathbb{R}$. $\square$

Now, consider the information $f(x_0-y_0)=f(y_0-x_0)$. We claim that $f(x_0-y_0)=f(y_0-x_0)\neq 0$.

Assume $f(x_0-y_0)=f(y_0-x_0)=0$.

$P(x_0,y_0) \implies f(x_0)^2 < f(y_0)^2$
$P(y_0,x_0) \implies f(x_0) \ge f(y_0)^2$
Contradiction. Therefore $f(x_0-y_0)=f(y_0-x_0)\neq 0$.

Therefore, there exist $a=x_0-y_0$ such that $f(a)=f(-a)\neq 0$. Using Claim $2$, we are finished. Done. $\blacksquare$
This post has been edited 5 times. Last edited by crocodilepradita, Jul 18, 2024, 12:59 AM
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YaoAOPS
1486 posts
#14
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Guess who didn't know what strict meant.


Note that if $f$ is a solution then $c \cdot f$ is a solution. As such, WLOG scale such that $f(1) = 1$.
Denote the assertion with $P(x, y)$. By $P(x, x)$ we get that $f(2x)f(0) \ge 0$ which finishes if $f(0) \ne 0$, so assume $f(0) = 0$.
Now, by $P(x, -x)$ we get that \[ 0 \ge f(x)^2 - f(-x)^2 \]Comparing with $P(-x, x)$, we get that $f(x)^2 = f(-x)^2$.

Claim: $f$ is either even or odd.
Proof. Suppose that $f(a) = f(-a) \ne 0, f(b) = -f(-b) \ne 0$. Then set $x + y = a, x - y = b$. Now $P(x, y), P(-x, -y)$ have LHS with opposite signs so $f(x)^2 - f(y)^2 \le 0$. By comparing $P(y, x), P(-y, -x)$ we get $f(y)^2 - f(x)^2 \le 0$ as well.
As such, $f(x)^2 = f(y)^2$ and thus $f(a)f(b) \ge 0, f(-a)f(-b) \ge 0$, contradiction. $\blacksquare$
First suppose $f$ is odd. Note that by $P(y, x)$ we have that $f(x + y)f(y - x) \ge f(y)^2 - f(x)^2$. Since $f(x + y)f(y - x) = -f(x + y)f(y - x) \le -(f(x)^2 - f(y)^2)$ it follows that equality must hold for all $x, y$, contradiction.
Thus, $f$ is even. Take $x_0, y_0$ such that $f(x_0 + y_0)f(x_0 - y_0) > f(x_0)^2 - f(y_0)^2$.
Now, suppose $f(a)$ has a negative sign. Then take $x + y = a, x - y = 1$.
We then get that \[ f(a) \ge f(x)^2 - f(y)^2 \]By $P(y, x)$, we also get \[ f(a) \le f(x)^2 - f(y)^2 \]This gives a contradiction.
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ericxyzhu
49 posts
#15 • 2 Y
Y by BlazingMuddy, Anzoteh
Anzoteh wrote:
Unfortunately I can't further characterize such functions. (What happens if we have $f(x + y) f(x - y) = f(x)^2 - f(y)^2$ and $f(x_0), f(y_0)$ different signs? How about the case where $f(x + y) f(x - y) > f(x)^2 - f(y)^2$ for all $x, y$? Those characterization seems elusive. )

The problem with the sign being equal was actually already an existing problem from IMOR 2018, proposed by Nuno Arala and Miguel Moreira from Portugal. https://artofproblemsolving.com/community/c6h1673292p10651933
And yes there are unexpected extraneous solutions.
This post has been edited 2 times. Last edited by ericxyzhu, Jul 18, 2024, 5:07 AM
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megarnie
5533 posts
#17
Y by
Let $P(x,y)$ denote the given assertion. Suppose that the result was not true and choose reals $a,b$ such that $f(a) < 0$ and $f(b) > 0$. Additionally, choose $c,d$ such that $f(c+d) f(c-d) > f(c)^2 - f(d)^2$.

Claim: $f(0) = 0$
Proof: Suppose otherwise.

$P(x,x): f(2x) f(0) \ge 0$. However, choosing $x = \frac{a}{2}$ implies that $f(a) f(0) \ge 0$, so $f(0) < 0$, but $x = \frac b2$ implies that $f(b) f(0) \ge 0$, so $f(0) > 0$, absurd. Thus, $f(0)$ must equal $0$. $\square$


$P(x,-x): 0 \ge f(x)^2 - f(-x)^2$. But $P(-x,x)$ gives $0\ge f(-x)^2 - f(x)^2$, so this means $f(x)^2 = f(-x)^2$, so either $f(x) = f(-x)$ or $f(x) = -f(-x)$.

Claim: $f(c-d) \ne -f(d-c)$
Proof: Suppose that $f(c-d) = -f(d-c)$. Notice that $f(c+d) f(c-d) > f(c)^2 - f(d)^2$ and $P(d, c)$ gives that $f(c+d)f(d-c) \ge f(d)^2 - f(c)^2$. Now, taking the negative of both sides and flipping the inequality gives that $f(c+d) f(c-d) \le f(c)^2  - f(d)^2$, absurd. $\square$

This implies that $f(c-d) = f(d-c)$.

$P \left( \frac{x + c - d}{2}, \frac{x - (c - d) }{2} \right): f(x) f(c-d) \ge f \left( \frac{x + c - d}{2} \right)^2 - f\left( \frac{x - (c-d)}{2} \right)^2 $

$P \left( \frac{x - (c - d)}{2} , \frac{x + c + d}{2} \right): f(x) f(d-c) = f(x) f(c-d)  \ge f\left( \frac{x - (c-d)}{2} \right)^2  -  f \left( \frac{x + c - d}{2} \right)^2$.

This implies if $k = f \left( \frac{x + c - d}{2} \right)^2 - f\left( \frac{x - (c-d)}{2} \right)^2$, then $f(x) f(c-d) \ge k$ and $f(x) f(c-d) \ge -k$, so $f(x) f(c-d) \ge 0$. However, if $f(c-d) = 0$, then $f(d-c) = 0$, contradiction to our claim that $f(c-d) \ne -f(d-c)$. Therefore, $f(c-d) \ne 0$, choosing $x\in \{a,b\}$ such that $f(x)$ and $f(c-d)$ have opposite signs gives that $f(x) f(c-d) < 0$, contradiction.

Hence we must have $f(x) \ge 0$ always or $f(x) \le 0$ always.
This post has been edited 2 times. Last edited by megarnie, Jul 19, 2024, 10:14 PM
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VicKmath7
1385 posts
#18
Y by
Solution
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SomeonesPenguin
123 posts
#19 • 1 Y
Y by zzSpartan
Interesting FE. :roll:

Solution
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SatisfiedMagma
450 posts
#20 • 1 Y
Y by Funcshun840
Okay retirement days are kinda interesting. I thought I forgot to flip the sign of the inequality somewhere and fakesolved. Turns out I was right all along! Here's a solution which should be right I think.

Solution: Denote $P(x,y)$ as the assertion to the inequality. With the usual swapping trick, consider $P(x,y)$ and $P(y,x)$. This will yield the following two inequalities.
\begin{align*}
f(x+y)f(x-y) &\ge f(x)^2 - f(y)^2 \\
f(x+y)f(y-x) &\ge f(y)^2 - f(x)^2
\end{align*}Add them up and write $y = x + a$ for some arbitrary real $a$.
\[f(2x+a) \left(f(-a) + f(a) \right) \ge 0.\]If $f$ is not an odd function, then we can find some $a_0 \in \mathbb{R}$ such that $f(a_0) + f(-a_0) \ne 0$. Then dividing the above inequality by $\left(f(a_0) + f(-a_0)\right)$ with an appropriate change in sign of inequality, we would be done. Henceforth assume $f$ is an odd function.

Since $f$ is odd, $f(x-y) = -f(y-x)$. From the above aligned equations, we can now say that
\[f(x+y)f(x-y) = f(x)^2 - f(y)^2 \]for all $x,y \in \mathbb{R}$ which is the desired contradiction since there exists some $x_0,y_0 \in \mathbb{R}$ such that there is strict inequality in $P(x,y)$. This completes the solution. $\blacksquare$
This post has been edited 1 time. Last edited by SatisfiedMagma, Aug 21, 2024, 2:28 AM
Reason: simple sign mistake, fixed!
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sansgankrsngupta
127 posts
#21
Y by
OG, Let $P(x,y)$ denote the given assertion.
$P(x,x) \implies$ $f(2x)f(0) \geq 0.$ If $f(0)$ is not 0, then the problem statement is directly implied, Hence, suppose $f(0)=0$
$P(x,-x) \implies$ $(f(x))^2 \geq (f(-x))^2$, replacing $x$ by $-x$ $\implies$ $(f(x))^2 \leq (f(-x))^2$. Thus, $(f(x))^2 =(f(-x))^2$ $\implies$ $f(x)= f(-x) or -f(-x)$. Now, Note that the problem condition can be translated to $f(x)f(y) \geq (f(\frac{x+y}{2}))^2-(f(\frac{x-y}{2}))^2$ for every $x,y\in\mathbb{R}$
Claim: $f(x)=f(-x)$ for all real $x$
Assume a real number $z /neq 0$(as $f(0)=0$) exists such that $f(z)=-f(-z)$.
Choose $x$ and $y$ such that and $x-y=z$
$P(x,y) \implies$
This post has been edited 1 time. Last edited by sansgankrsngupta, Sep 13, 2024, 11:15 AM
Reason: -
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N3bula
253 posts
#22
Y by
Let $P(x, y)$ denote the assertion.
\[P(x_0, y_0)+P(y_0, x_0)\]\[f(x_0+y_0)(f(x_0-y_0)+f(y_0-x_0))>0\]Let $s=x_0-y_0$.
\[P(x, x+s)+P(x+s, x)\]\[f(2x+s)(f(s)+f(-s))\geq 0\]Thus for all $x$ we have the desired result as $(f(s)+f(-s))>0$.
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bobthesmartypants
4337 posts
#23
Y by
Clearly $f$ is not identically 0; since $f$ satisfies the FE iff $-f$ satisfies the FE, WLOG let $f(1)>0$, and denote the FE as $P(x, y)$.

$P(x/2, x/2)\implies f(x)f(0) \ge 0$, so either $f(x)\ge 0$ in which we're done, or $f(0)=0$.

$P(x, -x)\implies 0 \ge f(x)^2-f(-x)^2\implies f(x)^2\le f(-x)^2$. But $f(-x)^2\le f(x)^2$ as well so $|f(x)|=|f(-x)|\quad (*)$.

$P(x, y) + P(y, x)\implies f(x+y)(f(x-y) + f(y-x))\ge 0$, and if we choose such that $x+y=1$ and $x-y=z \in \mathbb{R}$, then $f(z)+f(-z)\ge 0$ for all $z\in \mathbb{R}$, which by $(*)$ implies $f(z)\ge 0$ or $-f(z)=f(-z)\ne 0$. Let the set of all $z$ that satisfy the latter be $N$; note that if $|N|=0$ we're done.

Note that if $x-y$ satisfies $-f(x-y)=f(y-x)$, then $f(x+y)f(x-y) = -f(x+y)f(y-x)\ge f(x)^2-f(y)^2\implies f(x+y)f(y-x)\le f(y)^2-f(x)^2$. But $P(y, x) \implies f(x+y)f(y-x)\ge f(y)^2-f(x)^2$ already so $f(x+y)f(y-x)=f(y)^2-f(x)^2$. But substituting $(x, y) \to (-x, -y)$, we similarly get $f(-x-y)f(x-y) = f(y)^2-f(x)^2\implies f(-x-y)f(y-x) = f(x)^2-f(y)^2\quad (**)$. If we choose to further specify that $x-y\in N$, adding with the previous equality, gives $$(f(x+y)+f(-x-y))f(y-x) = 0\implies -f(x+y)=f(-(x+y))$$as $f(y-x) \ne 0$. Thus, if $|N|>0$, then $f(x)=-f(x)$ for all $x\in \mathbb{R}$.

This finally gives a contradiction as this would imply, by $(**)$, that $f(-x-y)f(y-x)=f(x)^2-f(y)^2$ for all $x, y\in \mathbb{R}$; yet the problem statement says there should exist $x_0, y_0$ for which this is not true. Thus, $|N|=0$ and we conclude $f(z)\ge 0$ for all $z\in \mathbb{R}$. $\blacksquare$
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poirasss
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#26 • 1 Y
Y by bin_sherlo
Putting $y=x$ gives us $f(2x)f(0) \geq 0 $. This gives $f(0)=0$, if not, then for all $x$, the sign of $f(x)$ doesn't change.
Then setting $x=-y$ gives $0\geq f(-y)^2-f(y)^2$, changing $y$ with $ -y$ gives us $f(y)^2\geq f(-y)^2\geq f(y)^2$, which implies $|f(x)|=|f(-x)|$ and $f(x)^2=f(-x)^2$.
Then swapping $-x$ as $x$ gives $f(y-x)f(-x-y)\geq f(x)^2-f(y)^2$ .
Swapping $x$ and $y$ gives $f(x-y)f(-x-y)\geq f(y)^2-f(x)^2$ which means $f(x)^2-f(y)^2\geq -f(x-y)f(-x-y)$.

Assume $f(a)>0>f(b)$. Choose $x$ and $y$ as $x-y=a$, $x+y=c$, where c is any real number.
$f(x+y)f(x-y)\geq f(x)^2-f(y)^2\geq -f(x-y)f(-x-y)$
$\Rightarrow f(c)f(a) \geq -f(a)f(-c)$
$\Rightarrow f(c) \geq -f(-c)$ Putting $-c$ as $c$ gives $f(c) \geq -f(-c) \geq f(c)$ which implies $f(c)=-f(-c)$.
It is given that $f(x_0+y_0)f(x_0-y_0)>f(x_0)^2-f(y_0)^2$ , $f(y_0+x_0)f(y_0-x_0)\geq f(y_0)^2-f(x_0)^2$.
So, $f(x_0+y_0)f(x_0-y_0)>f(x_0)^2-f(y_0)^2 \geq f(x_0+y_0)f(x_0+y_0)$, which gives us contradiction. $\blacksquare$
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InterLoop
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#27
Y by
kawaii
solution
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HamstPan38825
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#28
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By setting $x=y$, $f(2x)f(0) \geq 0$. Thus, if $f(0) \neq 0$, we may assume without loss of generality that $f(0) > 0$, from where $f(2x) \geq 0$ for all real numbers $x$, so the desired result is true.

If $f(0) = 0$, then \[f(x)^2-f(-x)^2 \leq f(0)f(2x) = 0.\]In particular, swapping $x$ and $-x$, it follows that $f(x)^2 = f(-x)^2$. Now, summing the equations for $(x, y)$ and $(y, x)$,
\[f(x+y)(f(x-y) + f(y-x)) \geq f(x)^2-f(y)^2+f(y)^2-f(x)^2 = 0.\]In particular, there is $(x_0, y_0)$ such that this inequality is strict; i.e. for this pair $(x_0, y_0)$, it follows that $f(c) = f(-c) \neq 0$ where $c = x_0-y_0$. However, for all pairs $(x, x+c)$, it follows that
\[f(2x+c)(f(c)+f(c))) \geq 0\]by the same argument, hence if $f(c) > 0$, $f(2x+c) \geq 0$ for all $x$ and vice versa.
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AshAuktober
920 posts
#29
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\subsection{Sol sketch}
We prove the contrapositive, i. e. given $a, b$ with $f(a) > 0, f(b) < 0$, equality must always hold.
The main claims are:
\begin{enumerate}
\item $f(0) = 0$
\item $f(x)^2 = f(-x)^2$.
\end{enumerate}
From here, swap $x, y$ to get $$f(x+y)(f(x-y) + f(y-x)) \ge 0.$$If $f(x-y) = -f(y-x)$, we're done as equality holds and thus must hold in the original equation.
Else if $f(x-y) = f(y-x)$, choosing $x, y$ such that $x-y$ comes out to the value required and $x+ y = a$, we have $f(x-y) \ge 0$. But doing the same such that $x+y = b$, $f(x-y) \le 0$.Thus $f(x-y) = 0$, which would mean (yet again) that equality holds. Thus we're done,
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Mr.Sharkman
487 posts
#30
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Claim: $f(0) = 0$ or the problem is true.

Proof: Applying $P(x,x),$ we get
$$f(2x)f(0) \ge f(x)^{2}-f(x)^{2} =0,$$so if $f(0) \neq 0,$ then all of $f(2x)$ have the same sign, and we are done.

Because of this, we can assume that $f(0)= 0. $

Claim: For all $x,$ $f(x)^{2} = f(-x)^{2}.$

Proof: Notice that, by $P(-x, x),$ we get $$f(0)f(-2x) \ge f(-x)^{2}-f(x)^{2},$$so $f(x)^{2} \ge f(-x)^{2}.$ Also, $f(-x)^{2} \ge f(x)^{2},$ so $f(x)^{2} = f(-x)^{2},$ as desired.

Now, by applying $P(x,y), P(y,x),$ and adding these together, we get
$$f(x-y)f(x+y)+f(y-x)f(x+y)\ge 0 \implies (f(x-y)+f(y-x))f(x+y) \ge 0.$$and we will call this assertion $Q(x,y).$

Claim: If $f(a) = f(-a)$ for some $a,$ then we are done.

Proof: Notice that taking $Q\left(\frac{a+b}{2},\frac{b-a}{2}  \right)$ gives
$$(f(a)+f(-a))f(b)\ge 0 \implies f(a)f(b) \ge 0.$$Since $b$ was arbitrary, we are done. $\blacksquare$

So, now, we need to find such an $a.$

Claim: $a = x_{0}-y_{0}$ satisfies $f(a) = f(-a).$

Proof: Assume FTSOC that $f(a)+f(-a)=0.$ Then, $Q$ would be at an equality case. However, since $Q$ is the sum of two inequalities, one of which is $P(x_{0}, y_{0}),$ which is not an equality case, we are done. $\blacksquare$
Thus, we have completed the proof. $\blacksquare$
This post has been edited 1 time. Last edited by Mr.Sharkman, Jan 29, 2025, 2:41 AM
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dubabuba
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#31
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Call the assertion $P(x, y)$.

Summing up $P(x, y)$ and $P(y, x)$ gives us $f(x + y) (f(x - y) + f(y - x)) \ge 0$.
For any $a, b \in \mathbb{R}$, there exists $x, y \in \mathbb{R}$ such that $a = x + y$ and $b = x - y$.
Therefore, $f(a) (f(b) + f(-b)) \ge 0$ for any $ a, b \in \mathbb{R} ~(\star) $.

Denote $a_0 = x_0 + y_0, b_0 = x_0 - y_0$.
Since the given inequality is strict for $(x_0, y_0)$, $~(\star)$ is also strict for $(a_0, b_0)$.
In otherwords, there exists $a, b \in \mathbb{R}$ such that $f(a)(f(b) + f(-b)) > 0$.

Suppose that $f(a_0) > 0$. This also implies $f(b_0) + f(-b_0) > 0$.
Plugging $(x, b_0)$ in $~(\star)$ gives $f(x)(f(b_0) + f(-b_0)) \ge 0 \Longrightarrow \forall x \in \mathbb{R}: f(x) \ge 0$.
The case $f(a_0) < 0$ can be done in a similar way.
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pie854
243 posts
#32
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Putting $x=y$ we get that $f(2x)f(0)\geq 0$ and so we are done if $f(0)\neq 0$. Suppose $f(0)=0$. Putting $x=-y$ we get $f(y)^2\geq f(-y)^2$ and so, since this holds for all $y$, $f(y)^2=f(-y)^2$.

Suppose there exists some $t$ with $f(t)\neq -f(-t)$ (in particular, note that $f(t)\neq 0$). Putting $x=\frac{a+t}2$, $y=\frac{a-t}2$ we get $$f(a)f(t)\geq f\left (\frac{a+t}2\right)^2-f\left (\frac{a-t}2\right)^2$$and putting $x=\frac{a-t}2$, $y=\frac{a+t}2$ we get $$f(a)f(t)\geq f\left (\frac{a-t}2\right)^2-f\left (\frac{a+t}2\right)^2.$$Thus, $f(a)f(t)\geq 0$ for all $a$ and we're done.

Now suppose $f(x)=-f(-x)$ for all real $x$. Putting $(x,y) \mapsto (y,x)$ we get $$-f(x+y)f(x-y)=f(y+x)f(y-x)\geq f(y)^2-f(x)^2 \implies f(x+y)f(x-y)\leq f(x)^2-f(y)^2,$$thus $f(x+y)f(x-y)=f(x)^2-f(y)^2$ for all $x,y$. But this is a contradiction with the assumption that the inequality is strict for some $x_0,y_0$.
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Ilikeminecraft
293 posts
#33
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what

Rewrite as $f(a)f(b)\geq f\left(\frac{a + b}2\right)^2 + f\left(\frac{a - b}2\right)^2.$ Plug in $b\mapsto-b,$ and add to get $f(a)(f(b) + f(-b))\geq0.$ Since there exist $a_0, b_0$ such that the inequality is strict, we have $f(a_0) + f(-a_0)>0$ or $f(a_0) + f(-a_0) < 0.$ The two cases are the same, so we assume it is positive. Plug in $b = a_0$ and we get $f(a)\geq0.$
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Nari_Tom
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#34
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We have that: $f(x+y)f(x-y) \geq f(x)^2-f(y)^2$.

By swapping (x,y) we will get that: $f(x+y)f(y-x) \geq f(y)^2-f(x)^2$.

Combining these we get: $f(x+y)[f(x-y)+f(y-x)] \geq 0$. If we take $(x=\frac{a+b}{2}, y=\frac{a-b}{2})$, then last inequality implies $f(a)[f(b)+f(-b)] \geq 0$, for all $a,b \in \mathbb{R}$. If there exist $c \in \mathbb {R}$ such that $f(c)+f(-c) \neq 0$, then $f(a)$ have same sign with $f(c)+f(-c)$.

But there exist such $c$, since $f(x_0+y_0)[f(x_0-y_0)+f(y_0-x_0)]>0$.
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