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Inequalities

Prove that ∑cyc1a3+b3+abc≤1abc\sum_{\text{cyc}} \frac{1}{a^3 + b^3 + abc} \leq \frac{1}{abc} for positive reals.

I was able to reduce the problem to ∑cyc1x+y+1≤1\sum_{\text{cyc}} \frac{1}{x+y+1} \leq 1 where x=a3x = a^3 and so on. Not sure what to do now.
Original post by esrever
Prove that ∑cyc1a3+b3+abc≤1abc\sum_{\text{cyc}} \frac{1}{a^3 + b^3 + abc} \leq \frac{1}{abc} for positive reals.

I was able to reduce the problem to ∑cyc1x+y+1≤1\sum_{\text{cyc}} \frac{1}{x+y+1} \leq 1 where x=a3x = a^3 and so on. Not sure what to do now.


Can't comment on your original problem, but if I understand it correctly, the one you've reduced it to isn't true. E.g. let a=b=c=0.1, then LHS = 3/1.002 which is greater than 1. Post your working so far, if you'd like someone to check it.
(edited 6 years ago)

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