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 $a^2\leq \frac{b^2+c^2 + 2ab}{4}$.

What if $a=\sqrt{3}$ and $b=c=1$?
For the edited question: by noting that $2b^2+2c^2\geq(b+c)^2$ (you can prove by expanding out the RHS), we have
$$
\frac{1}{2}(b^2+c^2)\geq\frac{1}{4}(b+c)^2=\left(\frac{b+c}{2}\right)^2\geq a^2.
$$
The last inequality above uses $0\leq a\leq(b+c)/2$.
