Functional Inequality Implies Uniform Sign

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Source: 2023 ISL A2

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#1 h Jul 17, 2024, 12:00 PM • 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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#2 h Jul 17, 2024, 12:01 PM • 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

#3 h Jul 17, 2024, 12:03 PM • 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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#4 h Jul 17, 2024, 12:12 PM • 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
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#5 h Jul 17, 2024, 12:25 PM • 1 Y

Y by SilverBlaze_SY

Thanks to Sammy27 for pointing out the point-wise trap in my previous (fake-)solution :ninja:

: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:

:thumbup:

This post has been edited 7 times. Last edited by kamatadu, Jul 18, 2024, 2:15 PM

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#6 h Jul 17, 2024, 12:27 PM

Y by

kamatadu wrote:

Can someone clarify what the $\gg$ means? I have a solution that seems to be pretty weird :maybe:

:maybe:

. Will be posting soon.

Sorry, I meant $\geqslant$ , non-strict inequality.

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#7 h Jul 17, 2024, 1:06 PM

Y by

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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#8 h Jul 17, 2024, 1:44 PM • 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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#9 h Jul 17, 2024, 2:54 PM • 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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#10 h Jul 17, 2024, 5:49 PM

Y by

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

#11 h Jul 17, 2024, 11:14 PM

Y by

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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#12 h Jul 17, 2024, 11:42 PM • 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. $\blacksquare$

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crocodilepradita

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crocodilepradita

#13 h Jul 18, 2024, 12:55 AM

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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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#14 h Jul 18, 2024, 4:09 AM

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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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#15 h Jul 18, 2024, 5:07 AM • 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
Reason: Typo

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#17 h Jul 19, 2024, 10:13 PM

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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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#18 h Jul 21, 2024, 12:17 AM

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Solution

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SomeonesPenguin

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#19 h Aug 6, 2024, 12:30 PM • 1 Y

Y by zzSpartan

Interesting FE. :roll:

:roll:

Solution

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#20 h Aug 16, 2024, 3:10 PM • 2 Y

Y by Funcshun840, H_Taken

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

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#21 h Aug 31, 2024, 11:44 AM

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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

#22 h Sep 14, 2024, 7:17 AM

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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

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bobthesmartypants

#23 h Sep 21, 2024, 10:20 PM

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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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#26 h Oct 1, 2024, 4:30 PM • 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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#27 h Oct 28, 2024, 8:33 AM

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kawaii
solution

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#28 h Dec 8, 2024, 7:47 PM

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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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#29 h Jan 25, 2025, 7:20 AM

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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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#30 h Jan 29, 2025, 2:41 AM

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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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#31 h Feb 1, 2025, 10:49 AM

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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

#32 h Mar 5, 2025, 11:28 AM

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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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#33 h Mar 10, 2025, 11:29 PM

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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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#34 h Mar 17, 2025, 7:09 AM

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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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#35 h Apr 16, 2025, 2:10 PM

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For $x = y$ , we get $f(2x)f(0) \geq f(x)^2 - f(x)^2 = 0$ . If $f(0) \neq 0$ , this implies $f(x)$ and $f(0)$ have the same sign, which finishes the problem.
Otherwise, $f(0) = 0$ . Taking $y = -x$ , we get $f(0)f(2x) \geq f(x)^2 - f(-x)^2$ , from which $f(-x)^2 \geq f(x)^2$ . Replacing $x$ with $-x$ , we get $f(x)^2 \geq f(-x)^2$ , and from the two inequalities we get $f(-x)^2 = f(x)^2, \forall x \in \mathbb{R}$ .

This implies $f(-x) = \pm f(x)$ . Replacing $x$ and $y$ in the statement, we get $f(y+x)f(y-x) \geq f(y)^2 - f(x)^2$ , and by summing with $f(x+y)f(x-y) \geq f(x)^2 - f(y)^2$ , we have $f(x+y)(f(x-y) + f(y-x)) \geq.$ We know there exist $x_0, y_0$ for which the inequality in the statement is strict, which means there exist $x_0, y_0$ for which $f(x_0+y_0)(f(x_0-y_0) + f(y_0-x_0)) > 0$ , which implies $f(x_0 - y_0) = f(y_0 - x_0)$ , otherwise their sum is null. Let $d = x_0 - y_0$ . Then $f(x + y)f(d) \geq 0$ , for any $x+y$ with $|x - y| = d$ . Let $x = d + y$ , then $f(2y+d)f(d) \geq 0$ , so $f(d + 2y)$ and $f(d)$ have the same sign, which solves the problem, as $2y + d$ takes any real value.

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