Let $0 \leqslant a, b, c \leqslant 1$. Show that $\frac{a}{b c+1}+\frac{b}{a c+1}+\frac{c}{a b+1} \leqslant 2$.

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The function $f(a, b, c)$ on the left-hand side is convex in each variable, so it reaches its maximum at one of the 8 points $\left(\epsilon_{1}, \epsilon_{2}, \epsilon_{3}\right)$ with $\epsilon_{i} \in\{0,1\}$ (indeed, we use the fact that if a function $g:[u, v] \rightarrow \mathbb{R}$ is convex, then $g$ reaches its maximum at $u$ or $v$). We verify that the maximum corresponds to ( $1,1,0$ ) and is 2.