Triangle - length of the sides - proof a, b and c are the lengths of the sides of a triangle. Prove that 
$$a^2+b^2 \ge \frac{1}{2}c^2$$  
Let $\gamma$ be the angle between sides a and b. then: 
$$a^2 + b^2 - 2ab\cos(\gamma) = c^2$$ 
Hence we need to prove that 
$$a^2+b^2 \ge \frac{1}{2}c^2$$ 
$$2a^2+2b^2 \ge a^2+b^2 -2ab\cos(\gamma) $$
$$a^2+b^2+2ab\cos(\gamma) \ge0$$ 
Knowing that $\cos(\gamma) \ge-1$, we get $a^2+b^2+2ab\cos(\gamma) \ge a^2 +b^2 -2ab =(a-b)^2$ 
I have carried this problem so far but now I am stuck. How should I proceed? 

You're already there. Your last line is 
$$a^2+b^2+2ab\cos(\gamma) \geq (a-b)^2$$
And so you only have to note $(a-b)^2\geq 0$ to see that 
$$a^2+b^2+2ab\cos(\gamma) \geq 0$$
Note that we can make this inequality strict ($>$) since $\cos(\gamma)=-1$ can't happen, for one angle is then stretched, so we could've taken $\cos(\gamma)>-1$ rather than $\cos(\gamma)\ge-1$.
