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?
3 Answers
$\begingroup$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 $\cos(\gamma)\ge-1$.
$\endgroup$ $\begingroup$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 .$
$\endgroup$ $\begingroup$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.
$\endgroup$
