Proving $ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $ for $0Let $a,b,c$ be positive real numbers between $0$ and $1$ ,i.e., they lie in the closed interval $[0,1]$. Prove that :
$$ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $$

Let $f(a,b,c)=\sum\limits_{cyc}\frac{a}{b+c+1}+\prod\limits_{cyc}(1-a)$.
Hence, $\frac{\partial^2f}{\partial a^2}=\frac{2b}{(a+c+1)^3}+\frac{2c}{(a+b+1)^3}\geq0$,
which says that $f$ is a convex function of $a$, of $b$ and of $c$.
Thus, $$\max_{\{a,b,c\}\subset[0,1]}f=\max_{\{a,b,c\}\subset\{0,1\}}f=f(1,1,1)=1.$$
Done!
