USAJMO problem 3: Inequality

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5818 posts

#1 h Apr 24, 2012, 9:57 PM • 12 Y

Y by Amir Hossein, HWenslawski, megarnie, chessgocube, Lilathebee, son7, ImSh95, Lamboreghini, Adventure10, Mango247, ItsBesi, and 1 other user

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

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NewAlbionAcademy

910 posts

NewAlbionAcademy

#3 h Apr 24, 2012, 10:05 PM • 9 Y

Y by HWenslawski, chessgocube, Lilathebee, megarnie, son7, ImSh95, Lamboreghini, Adventure10, and 1 other user

0. >.<

My friend said you needed to prove

$\frac{a^3+3b^3}{5a+b} \ge \frac{1}{2}a^2+\frac{1}{6}b^2$ or something like that.

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Minamoto

233 posts

Minamoto

#4 h Apr 24, 2012, 10:06 PM • 11 Y

Y by chessgocube, Lilathebee, Flying-Man, HWenslawski, son7, ImSh95, fidgetboss_4000, Lamboreghini, aidan0626, Adventure10, and 1 other user

I showed that when $a = b = c$ we have LHS = RHS

Then I set $a+b+c = x$ and showed that as $c$ approaches $x$ and $a$ and $b$ approach $0$ , we have that $\frac{16}{5} \geq \frac{2}{3}$ Only, I'm not sure whether I wrote $c$ approaches $x$ or infinity...

Then I got clueless, so I guessed "It is easy to show that for any $a,b,c$ between these two extremes, LHS is still greater than or equal to RHS, QED.

AkshajK wrote:

How many points do you think i'll get?

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exmath89

2572 posts

exmath89

#5 h Apr 24, 2012, 10:08 PM • 7 Y

Y by chessgocube, Lilathebee, HWenslawski, son7, ImSh95, Adventure10, and 1 other user

I got the inequality to $\sum_{cyc} \frac{a^3+3b^3}{5a+b}\ge\frac{6abc}{a+b+c}$ using AM-GM, but just bsed the rest to show the desired. -_-

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BOGTRO

5818 posts

BOGTRO

#6 h Apr 24, 2012, 10:10 PM • 7 Y

Y by Lilathebee, chessgocube, HWenslawski, son7, ImSh95, Adventure10, and 1 other user

Neither of those looks like it'll get a positive score :3

So I wrote
"By rearrangement inequality on {a,b,c} and {a^2,b^2,c^2}, we have $a^3+b^3+c^3 \geq ab^2+bc^2+ca^2$ . Thus, $12a^3+12b^3+12c^2 \geq 10a^3+10b^3+10c^2+2(ab^2+bc^2+ca^2)$

Thus, $(a^3+3b^3)+(b^3+3c^3)+(c^3+3a^3) \geq \frac{2}{3}a^2(5a+b)+\frac{2}{3}b^2(5b+c)+\frac{2}{3}c^2(5c+a)$ "

And stopped. Obviously this doesn't do a whole lot (well, nothing). Any possibility of sneaking a point?

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applepi2000

2226 posts

applepi2000

#7 h Apr 24, 2012, 10:10 PM • 8 Y

Y by anser, Lilathebee, chessgocube, HWenslawski, son7, ImSh95, Adventure10, and 1 other user

Probably the intended solution:
Click to reveal hidden text
My solution:
Click to reveal hidden text

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colinhy

751 posts

colinhy

#8 h Apr 24, 2012, 10:13 PM • 12 Y

Y by RudraRockstar, centslordm, Lilathebee, chessgocube, HWenslawski, fidgetboss_4000, son7, ImSh95, Adventure10, Mango247, Sedro, and 1 other user

Well, what I did is that I didn't really know what to do after an hour of bashing and stuff, so I used Cauchy, used a faulty muirhead (flipped the inequality sign

), and proved that. Hopefully they won't notice... or read this post.

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v_Enhance

6877 posts

v_Enhance

#9 h Apr 24, 2012, 10:30 PM • 16 Y

Y by Wizard_32, Kanep, HamstPan38825, myh2910, megarnie, centslordm, Lilathebee, chessgocube, HWenslawski, son7, Jc426, ImSh95, NS2k7, Rounak_iitr, Adventure10, Sanjana42

My Solution
By the way, I didn't take the JMO, but this took me about 20 minutes. I guess I probably should have tried Cauchy first, as above...

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tc1729

1221 posts

tc1729

#10 h Apr 24, 2012, 10:32 PM • 7 Y

Y by Lilathebee, chessgocube, megarnie, HWenslawski, ImSh95, Adventure10, Mango247

Is there anyway to legitly solve this with Nesbitt's? because i saw the $\frac{2}{3}$ , flipped it and that was the motivation for my bs.

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aZpElr68Cb51U51qy9OM

1600 posts

aZpElr68Cb51U51qy9OM

#11 h Apr 24, 2012, 11:23 PM • 11 Y

Y by Lilathebee, chessgocube, HWenslawski, Invincible0123, ImSh95, Iora, megarnie, Adventure10, Mango247, ehuseyinyigit, Sedro

I just used Titu's Lemma and used that $a^2+b^2+c^2 \ge ab+bc+ca$ .

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zero.destroyer

813 posts

zero.destroyer

#12 h Apr 24, 2012, 11:53 PM • 7 Y

Y by megarnie, Lilathebee, chessgocube, HWenslawski, ImSh95, Adventure10, Mango247

How much credit would they give if I said there exists an x such that (a^3+3b^3)/(5a+b)>=x*a^2+(2/3-x)b^2, and this is shown when letting q=(a/b), we have x*f(q)>=g(q) for all positives q, where f, g are quartic polynomials, and since it's well known that there's always a global maximum for the ratio of two polynomials with the same degree, x exists? Basically, this seems like the motivation for a well known technique (which I didn't know), where you use weighted am-gm a bunch.

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somenub

53 posts

somenub

#13 h Apr 25, 2012, 12:01 AM • 61 Y

Y by nsun48, BILL9, explogabloger, fractals, hamup1, AKAL3, 62861, champion999, brianapa, blawho12, WhaleVomit, bguo, droid347, oceanair, Wizard_32, biomathematics, RudraRockstar, mira74, Kanep, Imayormaynotknowcalculus, OlympusHero, Martin007B, jacoporizzo, tigerzhang, electrovector, vsamc, megarnie, ASweatyAsianBoie, Lilathebee, lethan3, tenebrine, ChrisWren, Toinfinity, mathking999, eagles2018, HWenslawski, pog, Jc426, ImSh95, Mathdreams, tree_3, aidan0626, ostriches88, ihatemath123, Adventure10, Mango247, EpicBird08, Ritwin, Jack_w, Sedro, ehuseyinyigit, hgomamogh, and 9 other users

yea i multiplied it out and used muirhead's 6 times, but hey that's why you have 4 1/2 hours

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anchenyao

292 posts

anchenyao

#14 h Apr 25, 2012, 12:06 AM • 16 Y

Y by fractals, champion999, Wizard_32, myh2910, tigerzhang, Lilathebee, ChrisWren, HWenslawski, ImSh95, Adventure10, Mango247, Ritwin, ehuseyinyigit, and 3 other users

It shouldn't take more than 1 hour with Muirhead if you know what you're doing. This problem screams of Muirhead, considering that both sides have the same degree, and multiplying will get 1296 terms on each side.

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proglote

958 posts

proglote

#15 h Apr 25, 2012, 12:13 AM • 14 Y

Y by champion999, OlympusHero, myh2910, Lilathebee, HWenslawski, ImSh95, Iora, Adventure10, Mango247, ehuseyinyigit, and 4 other users

lol solved this in a minute literally.. but I didn't take the test :

just WLOG $a^2 + b^2 + c^2 = 3$ and use CS: ${\sum \frac{a^4}{5a^2 + ab} + 3 \sum \frac{b^4}{5ab + b^2} \ge \frac{9}{15 + ab+bc+ca} + \frac{27}{5(ab+bc+ca)+3} \ge 2 \iff ab+bc+ca \le 3._{\blacksquare}}$

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somenub

53 posts

somenub

#16 h Apr 25, 2012, 12:13 AM • 6 Y

Y by nsun48, Lilathebee, HWenslawski, ImSh95, Adventure10, Mango247

2 muirheads could have been enough but more doesn't hurt, i still failed to solve problem 1 in the 2 1/2 hours i spent on it cause im terrible at geo

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trk08

614 posts

trk08

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

Y by

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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joshualiu315

2534 posts

joshualiu315

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

Y by

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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pqr.

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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bjump

1032 posts

bjump

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

Y by

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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Math4Life7

1703 posts

Math4Life7

#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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eg4334

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.

11 posts

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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little-fermat

147 posts

little-fermat

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

Y by

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

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cursed_tangent1434

639 posts

cursed_tangent1434

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

Y by

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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reni_wee

52 posts

reni_wee

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

Y by

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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megahertz13

3184 posts

megahertz13

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

Y by

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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Marcus_Zhang

980 posts

Marcus_Zhang

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

Y by

Bashing :P

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

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Ilikeminecraft

658 posts

Ilikeminecraft

#114 h Apr 25, 2025, 6:56 AM • 1 Y

Y by MihaiT

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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justaguy_69

14 posts

justaguy_69

#115 h May 5, 2025, 4:00 AM

Y by

seperate the numerators and titu

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Adywastaken

51 posts

Adywastaken

#116 h May 20, 2025, 5:17 PM

Y by

Notice
$\sum_{cyc} \frac{a^4}{5a^2+ab}\ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+ab+bc+ca}\ge \frac{a^2+b^2+c^2}{6}$ And
$3\sum_{cyc} \frac{b^4}{5ab+b^2}\ge \frac{(3(a^2+b^2+c^2))^2}{15(ab+bc+ca)+3(a^2+b^2+c^2)}\ge \frac{a^2+b^2+c^2}{2}$ Adding gives the result.

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