Prove $a^2+b^2\geq \frac{c^2}{2}$ and friends if $a+b\geq c\geq0$ Sorry for my inequality spam, but I got to prepare for my exams today :( Here's another:
Problem:

Prove$$a^2+b^2\geq \frac{c^2}{2}$$$$a^4+b^4\geq \frac{c^4}{8}$$$$a^8+b^8\geq \frac{c^8}{128}$$ if $a+b\geq c\geq0$

Attempt:
Working backwards:
$$a^2+b^2\geq \frac{c^2}{2}$$
$$\implies 2a^2+2b^2\geq c^2$$
I am stuck on the first one, let alone the others. I know by AM-GM, $\frac{a^2+b^2}{2}\geq ab$ but how can I use it here?

$(a^2+b^2)+(2ab)\ge c^2\implies $ at least one of the two terms in the LHS must be $\ge c^2/2$. Evidently this must be the larger one, viz $(a^2+b^2)$. 
Now the second and third inequalities follow from the first (replace $a,b,c$ with $a^2, b^2, c^2/2$ etc). 
