If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq (b^2 + c^2)/2$? If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq \frac{b^2 + c^2}{2}$? I can only see how to get $a^2\leq \frac{b^2+c^2 + 2ab}{4}$.

Let $b=M-d$ and $c=M+d$ then from $ a \le {M-d + M+d \over 2 } $ we get the reformulation $$ \tag 1 a \le M $$
Now take the squares
$$\tag{lhs} a  \to  a^2 $$
$$ \tag{rhs} {(M-d)^2 + (M+d)^2 \over 2 } = M^2+ d^2  $$
So if $ a \le M $ as assumed in $(1)$ then surely $$ \tag 2 a \le M = {b+c \over 2}\to a^2 \le M^2+d^2 = { b^2+c^2\over 2}$$
which answers the question in generality.
