Inequality with triangle sides

Question

Let $a,b,c$ be the sides of a triangle. Show that: $$(\sqrt a + \sqrt b - \sqrt c)(\sqrt a - \sqrt b + \sqrt c)(-\sqrt a + \sqrt b + \sqrt c) \ge \sqrt {(a+b-c)(a-b+c)(-a+b+c)}.$$

Answer 1

Let $\sqrt{a}=x$, $\sqrt{b}=y$ and $\sqrt{c}=y$.

Hence, we need to prove that $\prod\limits_{cyc}(x+y-z)^2\geq\prod\limits_{cyc}(x^2+y^2-z^2)$.

We'll prove that the last inequality is true for all reals $x$, $y$ and $z$.

Indeed, we can assume that $\prod\limits_{cyc}(x^2+y^2-z^2)>0$ and from here

we obtain $x^2+y^2-z^2>0$, $x^2+z^2-y^2>0$ and $y^2+z^2-x^2>0$ because

if for example, $x^2+y^2-z^2<0$ and $x^2+z^2-y^2<0$ so $x^2<0$, which is a contradiction.

Now easy to see that $$(x+y-z)^2(x+z-y)^2-(x^2+y^2-z^2)(x^2+z^2-y^2)=2(y^2+z^2-x^2)(y-z)^2\geq0$$

and we are done!