sahadian wrote: for real number $a,b,c$ in interval $ (0,1]$ prove that: $\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1} \leq 2$ Let $f(a,b,c)=\sum_{cyc}\frac{a}{bc+1}$. Hence, $f$ is a convex function of $a$, of $b$ and of $c$. Thus, $\max_{\{a,b,c\}\subset[0,1]}f=\max\{f(0,0,0),f(0,0,1),f(0,1,1),f(1,1,1)\}=f(0,1,1)=2$.

Or use $bc+1 \geq abc+1$ and $2+2abc-a-b-c=abc+(1-a)(1-bc)+(1-b)(1-c) \geq 0$.

again I see the question it is in mosp 2006 blue group the question is for interval $(0,1]$ and use $\leq $ but in this intervals we dont have any equality so I think they should use $<$

I think sahadian is right, the inequality should be strict. Note that if the variables $a,b,c$ could take the value $0$ (in other words if the interval $(0,1]$ is closed: $[0,1]$) then as Michael Rozenberg proved, the maximum of $f(a,b,c)$ is attained if $(a=0,b=1,c=1)$ and all such permutations. But $a>0$ so equality can't occur in this case.