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

The original question asked us to show that if $a < b+c$ then $a^2 < b^2+c^2$ which is why you can see a lot of answers (including mine) that say it is false:
It isn't true in general! If $a=5,\, b=2$ and $c=4$ then:
$$5<2+4$$
But $$25 \not\lt 4 + 16$$
Edit: Answering the updated question:
If $a \le \frac{b+c}{2} $ then $$a^2 \le \left(\frac{b+c}{2}\right)^2.$$ And since $$\frac{b^2+c^2}{2} \ge \left(\frac{b+c}{2}\right)^2,$$ we get that $$a^2 \le \frac{b^2+c^2}{2}.$$
As required.
It is not immediately obvious how to prove the inequality used in the middle of that argument, $\frac{b^2+c^2}{2} \ge \left(\frac{b+c}{2}\right)^2$. But if you expand it out you find: $$\frac{b^2+c^2}{2} \ge \frac{b^2+c^2+2bc}{4},$$ $$2b^2+2c^2 \ge b^2+c^2+2bc,$$ $$b^2+c^2 \ge 2bc \qquad\mbox{for b, c > 0}.$$ Which can be seen to be true by rewriting $c$ as $b+\alpha$ for some $\alpha \in \mathbb{R}$ and working the algebra through:
$$b^2 + (b+\alpha)^2 \ge 2b(b+\alpha),$$
$$2b^2 + \alpha^2 + 2b\alpha \ge 2b^2 + 2b\alpha,$$
$$\alpha^2 \ge 0.$$ Which is true for all $\alpha$.
