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Inequality in triangle
Nguyenhuyen_AG   0
23 minutes ago
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
0 replies
Nguyenhuyen_AG
23 minutes ago
0 replies
Reflected point lies on radical axis
Mahdi_Mashayekhi   5
N an hour ago by Mahdi_Mashayekhi
Source: Iran 2025 second round P4
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
5 replies
Mahdi_Mashayekhi
Apr 19, 2025
Mahdi_Mashayekhi
an hour ago
Find the value
sqing   18
N an hour ago by Yiyj
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
18 replies
sqing
Jun 22, 2024
Yiyj
an hour ago
Number Theory
fasttrust_12-mn   14
N 2 hours ago by Namisgood
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
14 replies
fasttrust_12-mn
Aug 15, 2024
Namisgood
2 hours ago
find question
mathematical-forest   5
N 2 hours ago by Jupiterballs
Are there any contest questions that seem simple but are actually difficult? :-D
5 replies
mathematical-forest
Thursday at 10:19 AM
Jupiterballs
2 hours ago
Own made functional equation
Primeniyazidayi   10
N 2 hours ago by Phat_23000245
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
10 replies
Primeniyazidayi
May 26, 2025
Phat_23000245
2 hours ago
Tough inequality
TUAN2k8   4
N 3 hours ago by Phat_23000245
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
4 replies
TUAN2k8
May 28, 2025
Phat_23000245
3 hours ago
Guess period of function
a1267ab   9
N 3 hours ago by HamstPan38825
Source: USA TST 2025
Let $n$ be a positive integer. Ana and Banana play a game. Banana thinks of a function $f\colon\mathbb{Z}\to\mathbb{Z}$ and a prime number $p$. He tells Ana that $f$ is nonconstant, $p<100$, and $f(x+p)=f(x)$ for all integers $x$. Ana's goal is to determine the value of $p$. She writes down $n$ integers $x_1,\dots,x_n$. After seeing this list, Banana writes down $f(x_1),\dots,f(x_n)$ in order. Ana wins if she can determine the value of $p$ from this information. Find the smallest value of $n$ for which Ana has a winning strategy.

Anthony Wang
9 replies
a1267ab
Dec 14, 2024
HamstPan38825
3 hours ago
Inequality with abc=1
tenplusten   11
N 3 hours ago by sqing
Source: JBMO 2011 Shortlist A7
$\boxed{\text{A7}}$ Let $a,b,c$ be positive reals such that $abc=1$.Prove the inequality $\sum\frac{2a^2+\frac{1}{a}}{b+\frac{1}{a}+1}\geq 3$
11 replies
tenplusten
May 15, 2016
sqing
3 hours ago
Central sequences
EeEeRUT   13
N 4 hours ago by v_Enhance
Source: EGMO 2025 P2
An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
13 replies
EeEeRUT
Apr 16, 2025
v_Enhance
4 hours ago
Interesting inequality
sqing   0
4 hours ago
Source: Own
Let $ a,b,c\geq  0 , a^2+b^2+c^2 =3.$ Prove that
$$ a^4+ b^4+c^4+6abc\leq9$$$$ a^3+ b^3+  c^3+3( \sqrt{3}-1)abc\leq 3\sqrt 3$$
0 replies
sqing
4 hours ago
0 replies
Advanced topics in Inequalities
va2010   23
N Apr 22, 2025 by Novmath
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
23 replies
va2010
Mar 7, 2015
Novmath
Apr 22, 2025
Advanced topics in Inequalities
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G H BBookmark kLocked kLocked NReply
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va2010
1276 posts
#1 • 11 Y
Y by DrMath, jh235, TheCrafter, Eugenis, mathleticguyyy, somebodyyouusedtoknow, Adventure10, Mango247, NicoN9, cubres, ehuseyinyigit
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
Attachments:
advanced-topics-inequalities (8).pdf (139kb)
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tastymath75025
3223 posts
#2 • 1 Y
Y by Adventure10
Will edit later maybe

4.1: sol

5.1: sol

[/hide]
This post has been edited 2 times. Last edited by tastymath75025, Mar 7, 2015, 5:31 AM
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Nuran2010
99 posts
#3 • 2 Y
Y by TunarHasanzade, TDVOLIMPTEAM
3.1:Click to reveal hidden text
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Nuran2010
99 posts
#4 • 3 Y
Y by TunarHasanzade, Strangett, TDVOLIMPTEAM
Similar logic,but a bit different approach for 5.1:
Click to reveal hidden text
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sqing
42518 posts
#5 • 1 Y
Y by Strangett
Problem 5.1. (Tuan Le)
Let $a,b,c$ be positive real numbers such that $  abc\geq 1$ Prove the inequality$$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\le 1.$$
Problem4.2. (T.Q.Anh)
Let $a,b,c>0$ and $ab+bc+ca=3$.Prove that$$ \frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2} \leq 1$$
Problem 3.1. (Titu Andresscu)
Let $ a,b,c>0 $ and $ a+b+c\geq 3 .$ Prove that$$ \frac{1}{a^2+b+c}+\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b}\leq 1$$https://artofproblemsolving.com/community/c6h1838005p12333824
This post has been edited 2 times. Last edited by sqing, Apr 16, 2025, 2:33 PM
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giangtruong13
152 posts
#6
Y by
You see NOTHING
This post has been edited 1 time. Last edited by giangtruong13, Apr 16, 2025, 1:55 PM
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Primeniyazidayi
117 posts
#7
Y by
The direction of 3.2 is false.Anyway,here is an approach from "Problems from the book,Titu Andreescu and Gabriel Dospinescu"
S3.2
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sqing
42518 posts
#8
Y by
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$
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sqing
42518 posts
#9
Y by
Let $x,y,z>0$ and $xyz=x+y+z$.Prove that$$\frac{(x-1)^2}{x^2+1}+\frac{(y-1)^2}{y^2+1}+\frac{(z-1)^2}{z^2+1} \geq 3-\frac{3\sqrt 3}{2}$$*
Let $ a, b, c\geq -\frac{3}{4},a+b+c=1 .$ Prove that
$$\frac{(a-1)^{2}}{a^2+1 }+\frac{(b-1)^{2}}{b^2+1 }+\frac{(c-1)^{2}}{c^2+1 }\geq \frac{6}{5}$$Let $ a, b, c>0,abc\geq 8 .$ Prove that $$\dfrac{(a-1)^2}{a^2+2}+\dfrac{(b-1)^2}{b^2+2}+\dfrac{(c-1)^2}{c^2+2}\geq\frac{25}{38}$$
This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 3:27 PM
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Strangett
2 posts
#10 • 1 Y
Y by Nuran2010
Problem 3.4. Given that \( a + b + c = 3 \), prove that
\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]
Click to reveal hidden text
This post has been edited 1 time. Last edited by Strangett, Apr 16, 2025, 5:18 PM
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Novmath
20 posts
#12 • 1 Y
Y by Strangett
Strangett wrote:
Problem 3.4. Given that \( a + b + c = 3 \), prove that
\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]
Click to reveal hidden text

Nice solution ,can you explain how did you think of taking f(x)= something with ln. When do we take f(x) as ln(something) ?
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Novmath
20 posts
#13
Y by
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

The problem 2.1 (USAMO 2001 P3) is wrong. The condition was $a^2+b^2+c^2+ abc =4$
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Strangett
2 posts
#14
Y by
Novmath wrote:
Strangett wrote:
Problem 3.4. Given that \( a + b + c = 3 \), prove that
\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]
Click to reveal hidden text

Nice solution ,can you explain how did you think of taking f(x)= something with ln. When do we take f(x) as ln(something) ?

Just in order to use the sum property of the natural logarithm (to apply Jensen's inequality to sums of function values). Also when taking the derivatives it is pretty obvious we are going to get something with its denominator squared and it is surely a convex function.
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sqing
42518 posts
#15
Y by
Novmath wrote:
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

The problem 2.1 (USAMO 2001 P3) is wrong. The condition was $a^2+b^2+c^2+ abc =4$
Yeah:
Let $a, b, c \geq 0$ and satisfy $ a^2+b^2+c^2 +abc = 4 . $ Show that\[ 0 \le ab + bc + ca - abc \leq 2. \]
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sqing
42518 posts
#16
Y by
Let $a, b, c \geq 0$ and $ a^2+b^2+c^2 +2abc = 4 . $ Show that$$ 0 \le ab + bc + ca - abc \leq 2$$
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Novmath
20 posts
#17
Y by
Strangett wrote:
Just in order to use the sum property of the natural logarithm (to apply Jensen's inequality to sums of function values). Also when taking the derivatives it is pretty obvious we are going to get something with its denominator squared and it is surely a convex function.
Thank you for explanation.
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Novmath
20 posts
#18
Y by
sqing wrote:
Novmath wrote:
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

The problem 2.1 (USAMO 2001 P3) is wrong. The condition was $a^2+b^2+c^2+ abc =4$
Yeah:
Let $a, b, c \geq 0$ and satisfy $ a^2+b^2+c^2 +abc = 4 . $ Show that\[ 0 \le ab + bc + ca - abc \leq 2. \]

Among a, b, c there must exist 2 numbers, say a and b, such that $ (1-a)(1-b) \geq 0$
Then: $ 1 \geq a + b - ab$
On the other hand, from $ a^2 + b^2 + c^2 + abc = 4$ we have:
$ c = \frac {-ab + \sqrt{a^2b^2 - 4a^2 - 4b^2 + 16}}{2} \leq \frac {-ab + \sqrt{a^2b^2 - 8ab + 16}}{2} = -ab +2$
Thus:
$ ab + bc + ca - abc = ab + c(a+b-ab) \leq ab + c \leq 2  (q.e.d)$
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Novmath
20 posts
#19
Y by
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
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sqing
42518 posts
#20
Y by
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474
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Novmath
20 posts
#21
Y by
sqing wrote:
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474

Thank you very much :coolspeak:
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Novmath
20 posts
#22
Y by
sqing wrote:
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474

Bro where can I learn separation trick in inequalities ,do you have some source which explaines it detailed, if you send it to me it will be appreciated!
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Novmath
20 posts
#23
Y by
sqing wrote:
Let $a, b, c \geq 0$ and $ a^2+b^2+c^2 +2abc = 4 . $ Show that$$ 0 \le ab + bc + ca - abc \leq 2$$

Can you also share solution of this problem:
(Vasc) Prove that if \( a, b, c \geq 0 \) and
\[
x = a + \frac{1}{b}, \quad y = b + \frac{1}{c}, \quad z = c + \frac{1}{a},
\]then
\[
xy + yz + zx \geq 2 + x + y + z.
\]
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Primeniyazidayi
117 posts
#24
Y by
Solution 3.3
This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 21, 2025, 8:34 PM
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Novmath
20 posts
#25
Y by
sqing wrote:
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474

Please share the solution of Vasc one that I shared
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