Geometric inequality with a triangle

Question

The positive real numbers $x,y,z$ are the side lengths of a triangle iff $$x^2 + y^2 + z^2 < 2\sqrt{x^2y^2 + y^2z^2 + z^2x^2}$$

Answer 1

$$ (x^2+y^2+z^2)^2\lt4(x^2y^2+y^2z^2+z^2x^2)\\ \Updownarrow\\ x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2\lt0\\ \Updownarrow\\ (x-y)^2(x+y)^2+z^4-2z^2\left(x^2+y^2\right)\lt0\\ \Updownarrow\\ \left(z^2-(x-y)^2\right)\left(z^2-(x+y)^2\right)\lt0\\ \Updownarrow\\ (z-x+y)(z+x-y)(x+y-z)(z+x+y)\gt0 $$ Which is true if and only if the Triangle Inequality is true for all sides.

If the quantity is not positive, at least one of the sides fails.

The perimeter factor, $x+y+z$, must be positive.

If two of the "triangle" factors is not positive, then one of the sides is not positive; e.g. if $(z-x+y)\le0$ and $(z+x-y)\le0$, then $2z=(z-x+y)+(z+x-y)\le0$. Thus, only one of the triangle factors can be negative.

So if the quantity is positive, all the factors must be positive.

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Heron's Formula says that the area of a triangle is $$ \frac14\sqrt{(z-x+y)(z+x-y)(x+y-z)(z+x+y)} $$ which is $$ \frac14\sqrt{4(x^2y^2+y^2z^2+z^2x^2)-(x^2+y^2+z^2)^2} $$