<s>[INST] A beautiful triangle inequality Prove that in any triangle with side lengths $a, b, c$ the inequality:
$$(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})\geq \sqrt{(-a+b+c)(a-b+c)(a+b-c)}.$$
For demonstration, I tried to use substitutions Ravi or algebraic calculation but without success. Maybe someone has an idea of saving. Thank you!
 [/INST] Since $\sqrt{b}+\sqrt{c}-\sqrt{a}=\sqrt{b+c+2\sqrt{bc}}-\sqrt{a}>\sqrt{a}-\sqrt{a}=0$, we need to prove that 
$$(a+b-c)^2(a+c-b)^2(b+c-a)^2\geq(a^2+b^2-c^2)(a^2+c^2-b^2)(b^2+c^2-a^2)$$
We can assume that $\prod\limits_{cyc}(a^2+b^2-c^2)\geq0$ and from here we can assume that
$a^2+b^2-c^2\geq0$, $a^2+c^2-b^2\geq0$ and $b^2+c^2-a^2\geq0$.
Indeed, if $a^2+b^2-c^2<0$ and $a^2+c^2-b^2<0$ so $a^2<0$, which is contradiction.
Now our inequality follows from 
$$(a+b-c)^2(a+c-b)^2-(a^2+b^2-c^2)(a^2+c^2-b^2)=2(b-c)^2(b^2+c^2-a^2)$$
Done!
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