USAMO 1997 Problem 5

Problem 5 from the 1997 USAMO seemed easy. Do you think my proof is correct?

Let $s=a^3+b^3+c^3+abc$ and $f(x)=\frac{1}{s-x^3}$ . Since the terms in the inequality are homogeneous, assume without loss of generality that $a+b+c=\sqrt[3]{s}$ , such that $0\lt a,b,c\lt\sqrt[3]{s}$ . For all $0\lt x\lt\sqrt[3]{s}$ , $f(x)$ is convex by the second derivative test. We have $f(a)+f(b)+f(c)\le3f(\frac{a+b+c}{3})=\frac{81}{26s}$ by Jensen's Inequality. To prove that $\frac{81}{26(a^3+b^3+c^3+abc)}\le\frac{1}{abc}$ , we have $\frac{81}{26}\le\frac{a^3+b^3+c^3+abc}{abc}$ and $\frac{81}{26}-4\le0\le\frac{a^3+b^3+c^3-3abc}{abc}=\frac{(a+b+c)((a-b)^2+(a-c)^2+(b-c)^2)}{2abc}$

#OlympiadMath #Proofs #MathProblem #Math

Easy Math Editor

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Comments

God I wish I knew how to do problems like this...

Kenneth Chan - 7 years, 9 months ago

Answer You are absolutely right ...

Mohammad mostafa Oladzad - 7 years, 9 months ago

Yeah,no flaws.Great work.

Priyatam Roy - 7 years, 9 months ago

Can you explain the WLOG? In particular, if $a=b=c$ , what would the WLOG values be?

Calvin Lin Staff - 7 years, 9 months ago

Do you need to explain your choice of WLOG on these exams?

Well, if it holds for $\{a,b,c\}$ such that $a+b+c=\sqrt[3]{s}$ , we can show it holds for $\{at,bt,ct\}$ where $t$ is a positive real:

$\sum_{cyc}\frac{1}{(at)^3+(bt)^3+abct^3}\le\frac{1}{abct^3}$

Multiplying by $t^3$ gives the desired result. So, for a set $\{a',b',c'\}$ , let $p=a'+b'+c'$ . If we multiply both sides by $\frac{\sqrt[3]{s}}{p}$ , we have $\{at,bt,ct\}=\{a',b',c'\}$ with $t=\frac{p}{\sqrt[3]{s}}$ , so it holds for all positive real $\{a,b,c\}$ .

Cody Johnson - 7 years, 9 months ago

@Cody Johnson – Yes, you have to explain the WLOG, and make sure that it indeed is WLOG. Your current application is not standard or typical, and you should review how to apply it properly. The reason I'm pointing it out, is that WLOG does not hold, but you have instead restricted the possible sets.

Note that you defined the variable $s$ twice, and need both scaled equations to hold. You have not answered how to deal with the case that $a = b = c =1$ . What is the scaling value of $t$ and the corresponding value of $s$ ?

Calvin Lin Staff - 7 years, 9 months ago

@Calvin Lin – OH I SEE WHAT I DID WRONG. Because it should be a constant, not a variable, for WLOG.

Cody Johnson - 7 years, 9 months ago

It's a valid proof, but I would argue that the second derivative test isn't necessarily fair game on the USAMO.

Michael Lee - 7 years, 9 months ago

I am pointing out that this is not a valid proof. Review the WLOG argument, in which he actually introduces a restrictive condition.

The second derivative test (for convexity) is fair game on the USAMO. However, make sure that you write out the full statements, instead of just claiming that it works (as was done above).

Calvin Lin Staff - 7 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$