TITLE: Sides of a triangle (square roots)?
QUESTION [5 upvotes]: This is the exercise: Let $a,b,c\in \mathbb{R}^+$. Prove that the following propositions are equivalents:

$a,b,c$ are sides of a triangle.
$\sqrt{a},\sqrt{b},\sqrt{c}$ are sides of an acute triangle. 

I'd really appreciate your help in this exercise. :)

REPLY [3 votes]: Let $\sqrt{a},\sqrt{b},\sqrt{c}$ be the sides of a non-acute triangle.  There is some angle $\theta$ in the triangle of at least 90 degrees.  Without loss of generality let it be opposite the side of length $\sqrt{c}$.  Then we have, from the Law of Cosines, $c=a+b-2\sqrt{ab}\cos\theta$.  We know $\cos\theta\le 0$.  Hence $c\ge a+b$, implying $a,b,c$ cannot be a triangle.