Show $\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} \geq \sqrt{a^2+ac+c^2}$ [closed]

Can someone help me with this problem:

Let $a$, $b$, $c$ be positive numbers. Show that:
$\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} \geq \sqrt{a^2+ac+c^2}$

Solution 1:

Let $OA=a, OB=b, OC=c, \angle AOB=60^{\circ}, \angle COB=60^{\circ}$

Then $$AB+BC\ge AC\Leftrightarrow\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} \geq \sqrt{a^2+ac+c^2}$$

Solution 2:

As an algebraic alternative to @Roman83's beautiful geometric solution.

Squaring both sides and applying Cauchy-Schwarz inequality \begin{align*} \sqrt{(a^2-ab+b^2)(b^2-bc+c^2)} &= \sqrt{\left[\left(b-\frac{a}2\right)^2 + \frac{3a^2}{4}\right]\left[\left(b-\frac{c}{2}\right)^2 + \frac{3c^2}{4}\right]} \\ &\geq \left|\left(b-\frac{a}{2}\right)\left(b-\frac{c}{2}\right)\right| + \frac{3ac}{4} \end{align*} The problem is reduced to $\left|(2b-a)(2b-c)\right| \ge -(2b-a)(2b-c)$, which is trivial.