<s>[INST] Prove the following inequality without using differentiation Let $a,b,c$ be real numbers that satisfy $0\le a,b,c\le 1$. Show that 
$$\frac a{b+c+1} + \frac b{a+c+1}  + \frac c{a+b+1} + (1-a)(1-b)(1-c) \le 1.$$
I don't know where to start. Multiplying everything by the denominators creates extreme mess.
 [/INST] Let $f(a, b, c)$ denote the left hand side of the inequality. Since $$\frac{\partial^2}{\partial a^2}f=\frac{2b}{(a+c+1)^3}+\frac{2c}{(a+b+1)^3}\ge0$$we have that $f$ is convex in each of the three variables; hence,
the maximum must occur where $a, b, c \in \{0, 1\}$. Since $f$ is $1$ at each of these $8$ points,
the inequality follows.
</s>