<s>[INST] 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?
 [/INST] A Geometric Proof:
Take a point $O$ in the plane and consider three line segments (this can be done since $x,y,z$ are positive real numbers) $OA, OB, OC$ with $$|OA|=x,|OB|=z,|OC|=y$$ and $$\angle AOB=60^{\circ},\angle BOC=60^{\circ}$$ therefore $\angle AOC=120^{\circ}$. In the triangle $\Delta ABC$ we have $$AB+BC\geq AC\tag{1}$$ Again by cosine rule  $$AB=\sqrt{x^2-xz+z^2}\\BC=\sqrt{y^2-yz+z^2}\\AC=\sqrt{x^2+xy+y^2}$$ Hence we get the desired inequality.
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