proving an inequality of sides of a triangle with trigonometry.

Question

On a textbook I was working with there was this question: When $a,b,c$ are the three sides of a triangle, prove that $a^2+b^2\gt\frac{c^2}{2}$ Since $(a-b)^2\ge0$, $a^2+b^2 \ge 2ab$, and $\cos\theta\gt-1$ ($\theta = $ the angle opposite to c, it follows that $ a^2+b^2 \ge 2ab \gt -2ab \cos\theta $, so $2(a+b)^2\ge a^2+b^2+2ab\gt a^2+b^2-2ab\cos\theta$, and I got stuck from here. How am I supposed to use the law of cosine here?

Answer 1

We know the cosine rule, $c^2 = a^2 + b^2 - 2ab\cos\theta$.
$a^2+b^2\gt\frac{c^2}{2}$ implies, $ a^2+b^2 \gt -2ab\cos\theta$
transferring $2ab$ term into LHS. we get $\frac{a^2+b^2}{2ab} \gt -\cos\theta$
Here, $LHS\ge1$ and $-1\lt RHS\lt1$ $(0\lt\theta\lt\pi)$.
Hence, proved.