If $a,b,c$ are real numbers such that $0\le a,b,c\le 1$, then $$\frac a{1+bc}+\frac b{1+ac}+\frac c{1+ab}\le 2.$$
I did the following:
Take $f(x)=\frac {x^2}{x+p},\ (p=abc)$ which is convex for $0\le x\le 1$. Therefore $$\begin{align}\frac {f(a)+f(b)+f(c)}3\le f(\frac {a+b+c}3)\Rightarrow\frac {a^2}{a+abc}+\frac {b^2}{b+abc}+\frac {c^2}{c+abc}\le \frac {(a+b+c)^2}{(a+b+c)+3abc}.\end{align}$$
How do I prove that the RHS is less than or equal to 2?
Note that $a,b,c \in [0,1] \implies abc \le ab,bc,ca$.
Thus, $\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab} \le \frac{a}{1+abc}+\frac{b}{1+abc}+\frac{c}{1+abc} = \frac{a+b+c}{1+abc}$
Since, $\displaystyle a,b,c \in [0,1] \implies a(1-b)(1-c) \ge 0 \implies \sum\limits_{cyc} a(1-b)(1-c) \ge 0 \\ \implies a+b+c + 3abc \ge 2(ab+bc+ac)$
$$\begin{align} \implies 2(abc+1) &\ge 2abc + 2 + 2(ab+bc+ac) - 3abc - (a+b+c) \\ &= a+b+c + abc + 2(1-a)(1-b)(1-c) \\ &\ge a+b+c \end{align}$$
Thus, $\displaystyle \frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab} \le \frac{a+b+c}{1+abc} \le 2$