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: $(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). 
A: 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.
A: All of them follow from Holder's inequality: for all $a_{ij}>0$:
$$\prod_{i=1}^k\sum_{j=1}^m a_{ij}^k\ge \left(\sum_{j=1}^m\prod_{i=1}^k a_{ij}\right)^k$$
$k=2$ gives Cauchy-Schwarz inequality. In this case, let $k=8, m=2, a_{ij}=1$ for all $i\in\{1,2,\ldots,7\}, j\in\{1,2\}$ and $a_{81}=a, a_{82}=b$:
$$(1^8+1^8)^7(a^8+b^8)\ge (a+b)^8\ge c^8$$
Similarly for the other two. Though this is unneeded for your simple inequalities: $2\left(a^2+b^2\right)\ge a^2+b^2+2ab=(a+b)^2\ge c^2$ and then, like Macavity suggests, replace $a,b,c$ with $a^2,b^2, c^2/2$ or with $a^4,b^4,c^4/2^3$.
