If $a$, $b$, and $c$ are positive numbers, then $\sqrt{a^2+ab+b^2}+\sqrt{a^2+ac+c^2}+\sqrt{b^2+bc +c^2}\ge\sqrt{3}(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}).$

Question

Prove that if $a$, $b$, and $c$ are positive real numbers, then $$\sqrt{a^2 + ab + b^2} + \sqrt{a^2 + ac + c^2} + \sqrt{b^2 + bc + c^2} \ge \sqrt{3} (\sqrt{ab} + \sqrt{ac} + \sqrt{bc}).$$ When does equality occur?

Answer 1

By AM-GM, we have $\sqrt{a^2+ab+b^2}\geq \sqrt{2ab+ab}=\sqrt{3ab}$.

Similarly, $\sqrt{a^2+ac+c^2}\geq \sqrt{3ac}$, $\sqrt{b^2+bc+c^2}\geq \sqrt{3bc}$.

Add them together, we have $$\sqrt{a^2+ab+b^2}+\sqrt{a^2+ac+c^2}+\sqrt{b^2+bc+c^2}\geq \sqrt{3}(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})$$

Done!