Does $a \leq b + c $ imply $a^2 \leq (b+c)^2 + (b-c)^2$? Givens
$$
a \leq b + c 
$$
or 
$$
a^2 \leq b^2 + c^2+2bc 
$$
Can we prove that?: 
$$
a^2 \leq (b+c)^2 + (b-c)^2 = 2b^2 + 2c^2
$$

$$(b-c)^2\geq0$$
therefore $$b^2+c^2\geq 2bc$$
Hence $$2b^2+2c^2\geq b^2+c^2+2bc\geq a^2$$
But the problem is that the very first inequality $b+c\geq a$ doesn't tell us if $a$ is non-negative. It is possible that $b+c\geq a$ but $a$ is very negative like $a=-\sqrt{(b^2+c^2)+1}$, while $b,c\geq0$. Then $b+c\geq0\geq a$ but $a^2=b^2+c^2+1>b^2+c^2$.
