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? 

How about this:
Triangle inequality theorem: a + b > c .
The sum of the lengths of two sides of a triangle is greater than the length of the third side.
1) $(a + b)^2 > c^2$ ; or
2)$ a^2 + 2ab + b^2 > c^2$.
Now:
3)$ (a -b)^2 \ge 0$;  hence : $a^2 + b^2 \ge 2ab$.
Putting together:
$ 2a^2 + 2b^2 \ge a^2 + 2ab + b^2 \gt c^2 .$
