Sides of a triangle (square roots)? 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. :)
 A: 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.
A: Hint: Suppose that $a,b,c$ are the sides of a triangle. Without loss of generality we may assume that $c\ge a$ and $c\ge b$. 
Then $a,b,c$ are the sides of a triangle if and only if $a+b\ge c$.
First we show that $\sqrt{a},\sqrt{b},\sqrt{c}$ are the sides of a triangle. We need to show that $\sqrt{a}+\sqrt{b}\gt \sqrt{c}$. This is easy, for if $\sqrt{a}+\sqrt{b}\le \sqrt{c}$ then, squaring, we obtain $a+2\sqrt{ab}+b\le c$, which implies that $a+b\lt c$.
Note that $a+b\gt c$  implies that $(\sqrt{a})^2+(\sqrt{b})^2 \gt (\sqrt{c})^2$. And this is precisely the condition for the largest angle of the triangle with sides $\sqrt{a}$, $\sqrt{b}$, $\sqrt{c}$ to be less than a right angle. For details, use the Cosine Law. 
We leave the proof in the other direction to you. 
