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?

Using the information you supplied, you could also just square both sides of the original inequality to obtain  $a^2+2ab+b^2 \geq c^2$. Divide the inequality by 2 and subtract $ab$ to the other side.  Given the AM-GM, we know that $\frac{a^2+b^2}{2} \geq ab $. Adding this inequality to the previous one we obtain $2\frac {a^2+b^2}{2} \geq \frac {c^2}{2} +ab-ab $. Simplifying gives us our desired result of $a^2+b^2\geq \frac {c^2}{2}$. 
As macavity said, you obtain the rest of the inequalities by substituting $a^2$ for $a $, $b^2$ for $b $, and so on.
