USAJMO problem 3: Inequality

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jlacosta

May 1, 2025

0 replies

USAJMO problem 3: Inequality

BOGTRO 104

N 3 hours ago by justaguy_69

Let $a,b,c$ be positive real numbers. Prove that $\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \geq \frac{2}{3}(a^2+b^2+c^2)$ .

104 replies

BOGTRO

Apr 24, 2012

justaguy_69

3 hours ago

2025 Math and AI 4 Girls Competition: Win Up To $1,000!!!

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68 replies

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Jan 26, 2025

RainbowSquirrel53B

4 hours ago

Question about AMC 10

MathNerdRabbit103 15

N 4 hours ago by GallopingUnicorn45

Hi,

Can anybody predict a good score that I can get on the AMC 10 this November by only being good at counting and probability, number theory, and algebra? I know some geometry because I took it in school though, but it isn’t competition math so it probably doesn’t count.

Thanks.

15 replies

MathNerdRabbit103

May 2, 2025

GallopingUnicorn45

4 hours ago

Arithmetic Series and Common Differences

4everwise 6

N 5 hours ago by epl1

For each positive integer $k$ , let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$ . For example, $S_3$ is the sequence $1,4,7,10,...$ . For how many values of $k$ does $S_k$ contain the term $2005$ ?

6 replies

4everwise

Nov 10, 2005

epl1

5 hours ago

9 Did I make the right choice?

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If you were in 8th grade, would you rather go to MOP or mc nats? I chose to study the former more and got in so was wondering if that was valid given that I'll never make mc nats.

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Martin2001

Apr 29, 2025

happypi31415

5 hours ago

find number of elements in H

Darealzolt 0

6 hours ago

If $H$ is the set of positive real solutions to the system
$x^3 + y^3 + z^3 = x + y + z$ $x^2 + y^2 + z^2 = xyz$ then find the number of elements in $H$ .

0 replies

Darealzolt

6 hours ago

0 replies

old problem from an open contest

Darealzolt 0

6 hours ago

Given that $a, b \in \mathbb{R}$ satisfy
$a + \frac{1}{a + 2015} = b - 4030 + \frac{1}{b - 2015}$ and $|a - b| > 5000$ . Determine the value of
$\frac{ab}{2015} - a + b.$

0 replies

Darealzolt

6 hours ago

0 replies

f_n(x)=\sum sin(nx)/n

Urumqi 6

N Today at 1:04 AM by Urumqi

$F_n(x)=\sum_{k=1}^{n}\frac{\sin (kx)}{k}$ , prove that for all $x \in (0,\pi), F_n(x)>0$ .

Thanks.

6 replies

Urumqi

Yesterday at 2:13 AM

Urumqi

Today at 1:04 AM

Looking for users and developers

derekli 9

N Today at 12:57 AM by musicalpenguin

Guys I've been working on a web app that lets you grind high school lvl math. There's AMCs, AIME, BMT, HMMT, SMT etc. Also, it's infinite practice so you can keep grinding without worrying about finding new problems. Please consider helping me out by testing and also consider joining our developer team! :P :blush:

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derekli

Yesterday at 12:57 AM

musicalpenguin

Today at 12:57 AM

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vanstraelen 6

N Yesterday at 11:36 PM by Math-lover1

Given the points $O(0,0,0),A(1,0,0),B(\frac{1}{2},\frac{\sqrt{3}}{2},0)$
a) Determine the point $C$ , above the xy-plane, such that the pyramid $OABC$ is a regular tetrahedron.
b) Calculate the volume.
c) Calculate the radius of the inscribed sphere and the radius of the circumscribed sphere.

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vanstraelen

Yesterday at 3:23 PM

Math-lover1

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Ecrin_eren 5

N Yesterday at 10:19 PM by imbadatmath1233

Let n be a natural number and p be a prime number. How many different pairs (n, p) satisfy the equation:

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Ecrin_eren

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Is there a name for the point $P'$ with respect to a circle $\Gamma$ , a diameter $\ell$ , and a given point $P$ , such that $P'$ is the reflection of the $P$ -antipode about $\ell$ ? Equivalently, $P'$ is the the other intersection of $\Gamma$ and the line through $P$ parallel to $\ell$ .

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A Collection of Good Problems from my end

SomeonecoolLovesMaths 5

N Yesterday at 8:46 PM by Math-lover1

This is a collection of good problems and my respective attempts to solve them. I would like to encourage everyone to post their solutions to these problems, if any. This will not only help others verify theirs but also perhaps bring forward a different approach to the problem. I will constantly try to update the pool of questions.

The difficulty level of these questions vary from AMC 10 to AIME. (Although the main pool of questions were prepared as a mock test for IOQM over the years)

Problem 1

Problem 2

Problem 3

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SomeonecoolLovesMaths

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Math-lover1

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GCD of consecutive terms

nsato 33

N Yesterday at 7:38 PM by reni_wee

The numbers in the sequence 101, 104, 109, 116, $\dots$ are of the form $a_n = 100 + n^2$ , where $n = 1$ , 2, 3, $\dots$ . For each $n$ , let $d_n$ be the greatest common divisor of $a_n$ and $a_{n + 1}$ . Find the maximum value of $d_n$ as $n$ ranges through the positive integers.

33 replies

nsato

Mar 14, 2006

reni_wee

Yesterday at 7:38 PM

USAJMO problem 3: Inequality

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#100 h Oct 10, 2023, 3:32 PM

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$\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \ge \frac{2a^2+2b^2+2c^2}{3}$
$(\frac{a^3}{5a+b}+\frac{b^3}{5b+c}+\frac{c^3}{5c+a})+(\frac{3b^3}{5a+b}+\frac{3c^3}{5b+c}+\frac{3a^3}{5c+a}) \ge \frac{2a^2+2b^2+2c^2}{3}$
$(\frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ac})+(\frac{3b^4}{5ab+b^2}+\frac{3c^4}{5bc+c^2}+\frac{3a^4}{5ac+a^2}) \ge \frac{2a^2+2b^2+2c^2}{3}$
By Titu's lemma, $(\frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ac})+(\frac{3b^4}{5ab+b^2}+\frac{3c^4}{5bc+c^2}+\frac{3a^4}{5ac+a^2}) \ge (\frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ac})+(\frac{3(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5ab+5bc+5ac})$
$a^2+b^2+c^2 \ge ab+bc+ac$ because $(2,0,0) > (1,1,0)$ by muirhead
So $(\frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ac})+(\frac{3(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5ab+5bc+5ac}) \ge (\frac{(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2})+(\frac{3(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2})=\frac{2a^2+2b^2+2c^2}{3}$

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614 posts

trk08

#101 h Oct 15, 2023, 12:43 AM

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It suffices to show:
$\sum{\text{cyc}}\frac{a^3}{5a+b}+3\sum{\text{cyc}}\frac{b^3}{5a+b}\geq\frac{2}{3}(a^2+b^2+c^2)$
By Titu's Lemma, the LHS is greater than or equal to:
$\frac{(a^2+b^2+c^2)^2}{\sum_{\text{cyc}}5a^2+\sum_{\text{cyc}}ab}+\frac{3(a^2+b^2+c^2)^2}{\sum_{\text{cyc}}5ab+\sum_{\text{cyc}}a^2}.$ Thus, it suffices to show:
$\frac{a^2+b^2+c^2}{\sum_{\text{cyc}}5a^2+\sum_{\text{cyc}}ab}+\frac{3(a^2+b^2+c^2)}{\sum_{\text{cyc}}5ab+\sum_{\text{cyc}}a^2}\geq \frac{2}{3}.$
Applying Titu's Lemma again, this is greater than or equal to:
$\frac{(4a)^2+(4b)^2+(4c)^2}{\sum_{\text{cyc}}8a^2+\sum_{\text{cyc}}16ab}=\frac{2a^2+2b^2+2c^2}{\sum_{\text{cyc}}a^2+\sum_{\text{cyc}}2ab}.$
Expanding, it suffices to show:
$3\sum_{\text{sym}}a^2\geq\sum_{\text{sym}}a^2+2\sum_{\text{sym}}ab,$ $2\sum_{\text{sym}}a^2\geq 2\sum_{\text{sym}}ab.$ As $(2,0,0)\succ (1,1,0)$ , the above inequality is true by Murihead's inequality.

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2534 posts

#102 h Oct 24, 2023, 5:50 PM

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Notice

$\begin{align*} \sum_{\text{cyc}} \frac{a^3}{5a+b} &= \sum_{\text{cyc}} \frac{a^4}{5a^2+ab} \\ &\ge \frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ca} \\ & \ge \frac{a^2+b^2+c^2}{6} \end{align*}$
Similarly, we obtain

$\begin{align*} \sum_{\text{cyc}} \frac{3b^3}{5a+b} &= \sum_{\text{cyc}} \frac{3b^4}{5ab+b^2} \\ & \ge \frac{3(a^2+b^2+c^2)^2}{5ab+5bc+5ca+a^2+b^2+c^2} \\ & \ge \frac{a^2+b^2+c^2}{2} \end{align*}$
Adding gives the desired results. $\square$

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174 posts

pqr.

#103 h Nov 26, 2023, 5:16 AM

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By Titu,
$\begin{align*} \sum_{\text{cyc}} \left(\frac{a^4}{a(5a+b)}+\frac{3b^4}{b(5a+b)}\right) &\ge \frac{(a^2+b^2+c^2)^2}{\sum_{\text{cyc}} (5a^2+ab)}+\frac{3(a^2+b^2+c^2)^2}{\sum_{\text{cyc}} (5ab+b^2)}, \end{align*}$ so it suffices to show that
$\begin{align*} \frac{a^2+b^2+c^2}{\sum_{\text{cyc}} (5a^2+ab)}+\frac{3(a^2+b^2+c^2)}{\sum_{\text{cyc}} (5ab+b^2)} &\ge \frac23. \end{align*}$ However, since $a^2+b^2+c^2 \ge ab+bc+ca$ , the inequality becomes
$\begin{align*} \frac{a^2+b^2+c^2}{6(a^2+b^2+c^2)}+\frac{3(a^2+b^2+c^2)}{6(a^2+b^2+c^2)} &\ge \frac23, \end{align*}$ which is clear after simplification.

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1013 posts

bjump

#104 h Jan 24, 2024, 2:27 AM

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Inequalities are fun

.
By Titu's Lemma
$\left( \sum_{\text{cyc}} \frac{a^{3}}{5a^{2}+ab}+3 \cdot \frac{b^{4}}{5ab+b^{2}} \right) \geq \frac{16(a^{2}+b^{2}+c^{2})^{2}}{8(a^{2}+b^{2}+c^{2})+16(ab+ac+bc)}= \frac{2(a^{2}+b^{2}+c^{2})^{2}}{(a+b+c)^{2}}$ Now it would suffice to show
$\frac{2(a^{2}+b^{2}+c^{2})^{2}}{(a+b+c)^{2}} \geq \frac{2}{3}(a^2+b^2+c^2)$ Or
$3a^{2}+3b^{2}+3c^{2} \geq (a+b+c)^{2}$ Expansion and rearrangement gives it suffices to show:
$(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$ Which is true by the trivial inequality $\square$

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1703 posts

#105 h Feb 26, 2024, 1:48 AM • 1 Y

Y by bjump

We let $\frac{a^3+3b^3}{5a+b} = \frac{a^4}{a(5a+b)} + \frac{3b^4}{b(5a+b)}$ . Applying Titu's lemma we get $\sum_{\text{cyc}} \frac{a^4}{a(5a+b)} + \frac{3b^4}{b(5a+b)} \geq \frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+ac+bc} + \frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5ab+5ac+5bc}$ Thus we just need to prove that $\frac{(a^2+b^2+c^2)}{5a^2+5b^2+5c^2+ab+ac+bc} + \frac{(a^2+b^2+c^2)}{a^2+b^2+c^2+5ab+5ac+5bc} \geq \frac{2}{3}$ This is obvious since $a^2+b^2+c^2 \geq ab+ac+bc$ . $\blacksquare$

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637 posts

eg4334

#106 h Sep 6, 2024, 3:27 AM

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Split the fractions apart and apply Titu's on the two triples to reduce the inequality into: $\frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+(ab+bc+ac)} + \frac{3(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5(ab+bc+ac)} \geq RHS$ or by cancelling the ugly variable term on the RHS: $\frac{a^2+b^2+c^2}{5(a^2+b^2+c^2)+ab+bc+ac} + \frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+5(ab+bc+ac)} \geq \frac23$ Sub $a^2+b^2+c^2=x$ and $ab+bc+ac=y$ for brevity and simplift further: $\frac{x}{5x+y} + \frac{3x}{x+5y} \geq \frac23$ $38x^2 \geq 28xy+10y^2$ $38 \left( \frac{x}{y}\right)^2 \geq 28 \left( \frac{x}{y} \right) + 10$ so we have reduced the inequality into $x \geq y$ which is trivial.

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jc.

#107 h Sep 14, 2024, 8:43 AM • 1 Y

Y by alexanderhamilton124

We prove the inequality by spitting sum into two easier sums and then finishing by using holder's inequalty and the following am-gm application $a^2+b^2+c^2 \geq ab+bc+ca$
$(\sum_{cyc}\frac{a^3}{5a+b})(\sum_{cyc}a(5a+b)) \geq (a^2+b^2+c^2)^2$ $\sum_{cyc}\frac{a^3}{5a+b} \geq \frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ca}$ $\geq\frac{(a^2+b^2+c^2)^2}{6(a^2+b^2+c^2)} = \frac{a^2+b^2+c^2}{6}$ and similarly using holder's inequality and am-gm for the remaining sum we get
$\sum_{cyc}\frac{3b^3}{5a+b}\geq \frac{3(a^2+b^2+c^2)^2}{a^2+b^2+c^2+5ab+5bc+5ca}$ $\geq \frac{3(a^2+b^2+c^2)^2}{6(a^2+b^2+c^2)} = \frac{a^2+b^2+c^2}{2}$ and finallyusing the above two inequalities we get
$\sum_{cyc}\frac{a^3+3b^3}{5a+b} = \sum_{cyc}\frac{a^3}{5a+b} + \sum_{cyc}\frac{3b^3}{5a+b}$ $\geq (a^2+b^2+c^2)(\frac{1}{2}+\frac{1}{6} = \frac{2}{3}(a^2+b^2+c^2)$

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#108 h Sep 16, 2024, 7:32 PM

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I discussed this problem in my youtube channel (little fermat) Video in my inequalities tutorial playlist.

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cursed_tangent1434

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cursed_tangent1434

#109 h Sep 19, 2024, 3:43 PM

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Simply note that from Titu's Lemma we have,
$\frac{a^3}{5a+b}+\frac{b^3}{5b+c}+\frac{c^3}{5c+a} = \frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ca} \ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+ab+c+ca}$ However, by the Rearrangement inequality we know, $a^2+b^2+c^2\ge ab+bc+ca$ . Thus,
$\frac{a^3}{5a+b}+\frac{b^3}{5b+c}+\frac{c^3}{5c+a} \ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+ab+c+ca} \ge \frac{(a^2+b^2+c^2)^2}{6(a^2+b^2+c^2)}=\frac{1}{6}(a^2+b^2+c^2)$ A similar argument also shows,
$\frac{3b^3}{5a+b}+\frac{3c^3}{5b+c}+\frac{3a^3}{5c+a} \ge \frac{1}{2}(a^2+b^2+c^2)$ Summing these inequality now yields,
$\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a} \ge \frac{1}{6}(a^2+b^2+c^2)+\frac{1}{2}(a^2+b^2+c^2)=\frac{2}{3}(a^2+b^2+c^2)$ which finishes the proof.

This post has been edited 1 time. Last edited by cursed_tangent1434, Sep 19, 2024, 3:43 PM
Reason: latex

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#110 h Sep 22, 2024, 4:15 AM

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Note that: As $(2,0,0) \succ (1,1,0)$ by Muirhead's Inequality,
, $a^2 + b^2 + c^2 \geq ab + bc +ca$ We can split the R.H.S. of the original equation as follows
$\sum_{cyc} (\frac{a^4}{5a^2 + ab} + \frac{3b^4}{5ab + b^2})$ Applying Titu's Lemma separately ,
$\begin{aligned} \sum_{cyc} \frac{a^4}{5a^2 + ab} &\geq \frac{(a^4+b^4+c^4)^2}{5( a^2 + b^2 + c^2) + ab + bc + ca}\\ &\geq \frac{(a^2+b^2+c^2)^2}{6( a^2 + b^2 + c^2)}\\ &\geq \frac{(a^2+b^2+c^2)}{6} \end{aligned}$ $\begin{aligned} \sum_{cyc} \frac{3b^4}{5ab + b^2} &\geq \frac{3(a^4+b^4+c^4)^2}{a^2 + b^2 + c^2 + 5(ab + bc + ca)}\\ &\geq \frac{3(a^2+b^2+c^2)^2}{6( a^2 + b^2 + c^2)}\\ &\geq \frac{(a^2+b^2+c^2)}{2} \end{aligned}$ Summing up the 2 inequalities
$\sum_{cyc} (\frac{a^4}{5a^2 + ab} + \frac{3b^4}{5ab + b^2}) \geq \frac{2}{3} (a^2 + b^2 + c^2).$

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#112 h Nov 3, 2024, 2:15 PM

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We want to prove that $\frac{a^3}{5a+b}+\frac{b^3}{5b+c}+\frac{c^3}{5c+a}+3\cdot(\frac{a^3}{5c+a}+\frac{b^3}{5a+b}+\frac{c^3}{5b+c})\ge \frac{2}{3}(a^2+b^2+c^2).$ By Titu's Lemma, $\frac{a^4}{5a^2+ab}+\frac{b^4}{5b^2+bc}+\frac{c^4}{5c^2+ac}+3\cdot(\frac{a^4}{5ca+a^2}+\frac{b^4}{5ab+b^2}+\frac{c^4}{5bc+c^2})$ $\ge \frac{(a^2+b^2+c^2)^2}{5a^2+5b^2+5c^2+ab+bc+ca}+3\cdot\frac{(a^2+b^2+c^2)^2}{5ab+5bc+5ca+a^2+b^2+c^2}$ $\ge \frac{(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2}+3\cdot\frac{(a^2+b^2+c^2)^2}{6a^2+6b^2+6c^2}$ $=\frac{2}{3}(a^2+b^2+c^2).$

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#113 h Mar 21, 2025, 9:56 PM

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Bashing :P

This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 21, 2025, 9:59 PM

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#114 h Apr 25, 2025, 6:56 AM

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By Holder's:
$\left(\sum \frac{a^3}{5a + b} + \frac{3b^3}{5a + b}\right)\left(\sum(5a + b)a + \sum(5a + b)3b\right)\geq16(a^2 + b^2 + c^2)^2$ dividing by the non LHS term, we have $2\cdot\frac{a^2 + b^2 + c^2}{(a + b + c)^2}.$ Thus, it is enough to prove that $(a + b + c)^2 \leq 3(a^2 + b^2 + c^2).$ This is pretty obvious by expanding and then applying muirheads, or completing the square. Thus, we are done.

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#115 h 3 hours ago

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seperate the numerators and titu

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