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For real numbers $a,b,c \in [0,1]$ prove inequality $$\frac a{1+bc}+\frac b{1+ac}+\frac c{1+ab}+abc\le \frac52$$

I tried AM-GM, Buffalo way. I do not know how to solve this problem

Michael Rozenberg
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Roman83
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1 Answers1

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Let $f(a,b,c)=\sum\limits_{cyc}\frac{a}{1+bc}+abc$.

$\frac{\partial^2f}{\partial a^2}=\frac{2bc^2}{(1+ac)^3}+\frac{2cb^2}{(1+ab)^3}\geq0$, which says that $f$ is a convex function of $a$, of $b$ and of $c$.

Thus, $f$ gets a maximal value for an extremal values of variables.

Id est, $\max f=\max\{f(0,0,0), f(0,0,1), f(0,1,1), f(1,1,1)\}=\frac{5}{2}$.

Michael Rozenberg
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  • Why $\frac{\partial^2}{\partial a^2},\frac{\partial^2}{\partial b^2},\frac{\partial^2}{\partial c^2}>0$ ensure convexity? That is not a sufficient condition for the hessian matrix to be positive definite. – Jack D'Aurizio Jun 08 '16 at 11:43
  • @JackDAurizio The function may not be globally convex, but when two variables are fixed the resulting univariate function is convex. This is sufficient for what we need here – Ewan Delanoy Jun 08 '16 at 11:57