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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?

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    $\begingroup$ what will you do further? your proof is complete $\endgroup$ – Dr. Sonnhard Graubner Apr 16 '17 at 10:31
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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 $\cos(\gamma)\ge-1$.

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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 .$

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The squared length of the median through $C$ is exactly $\frac{1}{4}\left(2a^2+2b^2-c^2\right)$ and that gives $a^2+b^2\geq \frac{1}{2}c^2$ as a straightforward consequence.

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