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 \[\frac{81}{26}-4\le0\le\frac{a^3+b^3+c^3-3abc}{abc}=\frac{(a+b+c)((a-b)^2+(a-c)^2+(b-c)^2)}{2abc}\]

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TopNewestGod I wish I knew how to do problems like this... – Kenneth Chan · 3 years, 11 months ago

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– Mohammad mostafa Oladzad · 3 years, 11 months ago

Answer You are absolutely right ...Log in to reply

Yeah,no flaws.Great work. – Priyatam Roy · 3 years, 11 months ago

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– Calvin Lin Staff · 3 years, 11 months ago

Can you explain the WLOG? In particular, if \(a=b=c\), what would the WLOG values be?Log in to reply

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 · 3 years, 11 months ago

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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 · 3 years, 11 months ago

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– Cody Johnson · 3 years, 11 months ago

OH I SEE WHAT I DID WRONG. Because it should be a constant, not a variable, for WLOG.Log in to reply

It's a valid proof, but I would argue that the second derivative test isn't necessarily fair game on the USAMO. – Michael Lee · 3 years, 11 months ago

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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 · 3 years, 11 months ago

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