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?
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$\begingroup$ See for example math.stackexchange.com/q/3692837/42969 $\endgroup$– Martin RFeb 10 at 17:13
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$\begingroup$ Equality occurs when $a=b=c$ $\endgroup$– VasiliFeb 10 at 17:47
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1 Answer
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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!
