USAMO 1997 Problem 5

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

Let s=a3+b3+c3+abcs=a^3+b^3+c^3+abc and f(x)=1sx3f(x)=\frac{1}{s-x^3}. Since the terms in the inequality are homogeneous, assume without loss of generality that a+b+c=s3a+b+c=\sqrt[3]{s}, such that 0<a,b,c<s30\lt a,b,c\lt\sqrt[3]{s}. For all 0<x<s30\lt x\lt\sqrt[3]{s}, f(x)f(x) is convex by the second derivative test. We have f(a)+f(b)+f(c)3f(a+b+c3)=8126sf(a)+f(b)+f(c)\le3f(\frac{a+b+c}{3})=\frac{81}{26s} by Jensen's Inequality. To prove that 8126(a3+b3+c3+abc)1abc\frac{81}{26(a^3+b^3+c^3+abc)}\le\frac{1}{abc}, we have 8126a3+b3+c3+abcabc\frac{81}{26}\le\frac{a^3+b^3+c^3+abc}{abc} and 812640a3+b3+c33abcabc=(a+b+c)((ab)2+(ac)2+(bc)2)2abc\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}

Note by Cody Johnson
6 years, 1 month ago

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6 votes

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

Kenneth Chan - 6 years, 1 month ago

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Answer You are absolutely right ...

Mohammad mostafa Oladzad - 6 years, 1 month ago

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Yeah,no flaws.Great work.

Priyatam Roy - 6 years, 1 month ago

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

Calvin Lin Staff - 6 years, 1 month ago

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

Well, if it holds for {a,b,c}\{a,b,c\} such that a+b+c=s3a+b+c=\sqrt[3]{s}, we can show it holds for {at,bt,ct}\{at,bt,ct\} where tt is a positive real:

cyc1(at)3+(bt)3+abct31abct3\sum_{cyc}\frac{1}{(at)^3+(bt)^3+abct^3}\le\frac{1}{abct^3}

Multiplying by t3t^3 gives the desired result. So, for a set {a,b,c}\{a',b',c'\}, let p=a+b+cp=a'+b'+c'. If we multiply both sides by s3p\frac{\sqrt[3]{s}}{p}, we have {at,bt,ct}={a,b,c}\{at,bt,ct\}=\{a',b',c'\} with t=ps3t=\frac{p}{\sqrt[3]{s}}, so it holds for all positive real {a,b,c}\{a,b,c\}.

Cody Johnson - 6 years, 1 month ago

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@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 ss twice, and need both scaled equations to hold. You have not answered how to deal with the case that a=b=c=1 a = b = c =1 . What is the scaling value of tt and the corresponding value of ss?

Calvin Lin Staff - 6 years, 1 month ago

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

Cody Johnson - 6 years, 1 month ago

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

Michael Lee - 6 years, 1 month ago

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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 - 6 years, 1 month ago

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