毕达哥拉斯不等式（三角学的衍生）

Recall the Pythagorean inequality:

If a triangle is acute, and $a<b<c$, then $a^2+b^2>c^2$

If it is right, and $a<b<c$, then $a^2+b^2=c^2$

Now the law of cosines states that $c^2=a^2+b^2-2ab \cos C$.

Now what happens if C is acute?

Since $\cos C$ is positive for all acute C, we have $c^2 = a^2 + b^2 - 2ab \cos C < a^2+b^2$

Similarly, if C is obtuse, then $\cos C < 0$, so

$c^2 = a^2+b^2 - 2ab \cos C > a^2+b^2$