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I've been trying to prove the following inequality
Assume that a,b,c are positive reals s.t. $abc=1$ , prove that : $$S=\frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} $$

If you apply AM-GM directy you get :

$$ S \geq \frac{3}{\sqrt[3]{(a^n+(n-1)b^n)(b^n+(n-1)c^n)(c^n+(n-1)a^n)}} $$

But from here I get stuck ... I think there is probably a change of variable or another trick (convex function + Jensen).

We can apply AM-GM again to get this : $$\sqrt[3]{(a^n+(n-1)b^n)(b^n+(n-1)c^n)(c^n+(n-1)a^n)} \leq \frac{n}{3} \left(a^ n+ b^n+c^ n \right) $$

$$ S \geq \frac{9}{n(a^n+b^n+c^n)} $$

Unfortunately $ a^n+b^n+c^n \geq 3 $ by AM-GM so it doesnt help to conclude
Thanks for your help.

Bart Michels
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dyoann
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  • I also tried with lagrange multipliers, hoping to prove we need a=b=c=1 ( for which you notice the equality holds), but I got also stuck in heavy calculations. – dyoann Oct 09 '14 at 20:22
  • Answered here http://math.stackexchange.com/questions/963977/i-have-used-cauchy-and-jensen-it-is-not-helping-me-very-much-advice-on-solving/966829#966829 – Macavity Oct 10 '14 at 15:36
  • Very nice, may I ask how you came with the idea of spliting a^(n+2)/(a^n+(n-1)b^n) into a^2 - (n-1)a^2b^n/(a^n+(n-1)b^n) ? Just experience with this kind of problem :-) ? – dyoann Oct 10 '14 at 17:40
  • It's a reasonably often used method - think called reverse Cauchy technique – Macavity Oct 10 '14 at 18:20
  • It helps when using AM-GM (or others) gives you a upper bound when you need a lower one, or vice versa. – Macavity Oct 10 '14 at 19:02
  • Thanks, I was able to find a description of this method on some online book, Indeed nice trick ! – dyoann Oct 10 '14 at 21:12

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