Problem 1137

Since $\max(x, y) = \frac{x+y+|x-y|}{2}$, twice the expression
$$a \max(b, c) + b \max(c, a) + c \max(a, b)$$
is equal to
$$a(b+c) + b(c+a) + c(a+b) + a|b-c| + b|c-a| + c|a-b|.$$
Therefore,
$$\begin{aligned} & 2(a \max(b,c) + b \max(c,a) + c \max(a,b)) - 2(a^2 + b^2 + c^2) \\ &= 2 \sum_{\text{cyc}} ab + \sum_{\text{cyc}} a|b-c| - 2 \sum_{\text{cyc}} a^2 \\ &= \sum_{\text{cyc}} a|b-c| - \sum_{\text{cyc}} (b-c)^2 = \sum_{\text{cyc}} |b-c|(a - |b-c|) \ge 0, \end{aligned}$$
since, by the triangle inequality, $|b-c| < a$, $|c-a| < b$, $|a-b| < c$. Moreover, the inequality is strict unless $|b-c| = |c-a| = |a-b| = 0$, i.e., $a = b = c$.

Both sides of the inequality remain unchanged under any permutation of $a, b, c$. Therefore, we may assume $a \le b \le c$. Under this assumption, $a \max(b,c) + b \max(c,a) + c \max(a,b) = ac + bc + cb = ac + 2bc$. This means we need to prove that $a^2 + b^2 + c^2 \le ac + 2bc$ which is equivalent to
$$(c-b)^2 \le a(c-a).$$
As $a \le b$, we have $c - b \le c - a$. Because $b \le c$ and $c < a + b$, we also have $0 \le c - b < a$. Multiplying these inequalities gives the desired result.
To understand the case of equality note that, because of the strict inequality $c - b < a$, we can only have $(c-b)^2 = a(c-a)$ if $c-b = c-a = 0$, which is equivalent to $a = b = c$.