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. :)
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$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
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