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For positive $a,b,c$, show that $$a+b+c = 3abc \implies \frac1{a+b}+\frac1{b+c}+\frac1{c+a} \leq \frac32$$

This "should" be the easiest case of an earlier question:

If $a+b+c = 3abc$ and $\frac17 \leq k \leq 7$ prove $ \frac1{ka+b}+\frac1{kb+c}+\frac1{kc+a} \leq \frac3{k+1} $

And I am told that 3-variable inequality problems with constraints "come from" triangle geometry problems.

Yet I can't make progress on even this case.

There are a huge bunch of problems without answers on this StackExchange site, of the following nature:

  • They concern three positive variables.

  • They ask to prove that some cyclic sum and/or product of expressions involving the variables is $\leq$ (or $\geq$ ) some constant or other cyclic sum.

  • They usually involve some constraint which is expressed as an equality relation between cyclic sums and/or products and/or constants.

  • The inequality is saturated (that is, equality is achieved) at $a=b=c$; usually, with scaling, one can find an equivalent problem where the equality is achieved at $a=b=c=1$.

  • No satisfactory answers are present at the StackExchange question.

In many cases, the problem will have been suggested by @Michael Rozenberg. I would love to find some tool kit for attacking such problems. I used to be really good at these, but I suspect my brain is aging out of that status.

Community
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Mark Fischler
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1 Answers1

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Since $$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=3,$$ it follows that $$\frac{2}{\sqrt{ab}}+\frac{2}{\sqrt{bc}}+\frac{2}{\sqrt{ca}}\le 1+\frac{1}{ab}+1+\frac{1}{bc}+1+\frac{1}{ca}=6.$$ So: $$\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\le 3.$$ Now $$\frac{1}{a+b}\le \frac{1}{2\sqrt{ab}},$$ and so on gives us $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\le \frac32.$$

Quang Hoang
  • 15,854