Proving $\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$ algebraically 
The question is to prove that for any positive real numbers $x$, $y$ and $z$,
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$

So I decided to do some squaring on both sides and expanding:
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
$$x^2 -xz+z^2 +y^2-yz+z^2+2\sqrt{(x^2-xz+z^2)(y^2-yz+z^2)}  - x^2 - xy - y^2 \geq 0$$
$$2\sqrt{(x^2-xz + z^2)(y^2-yz+z^2)}\geq xz+yz+xy-2z^2$$
Squaring both sides yields
$$4(x^2-xz+z^2)(y^2-yz+z^2)\geq(xy+xz+yz-2z^2)^2$$
After some expanding it becomes
$$(xy-xz-yz)^2 \geq 0$$
which completes the proof.
However, I want to ask whether the squaring in the $3^{rd}$ - $4^{th}$step is problematic. Since it is possible for $xy + xz+ yz -2z^2$ to be negative. Squaring both sides will invert the sign. Do I have to analyse the positive and negative case here seperately?

Let there be a triangle $ABC$ with vertices as $A(x,0), B(z/2,z\sqrt{3}/2), C(-y/2,y\sqrt{3}/2),$ then $$AB=\sqrt{x^2+z^2-xz},~ BC=\sqrt{y^2+z^2-yz}, ~CA=\sqrt{x^2+y^2+xy}.$$
Finally, the triangular inequality  $$AB+BC \ge CA.$$
proves the result. The equality holds when $A,B,C$ are collinear.
