# Proof about an inequality

We have $a,b,c>0$, prove that:

$$\dfrac {1+\sqrt {3}} {3\sqrt {3}}\left( a^{2}+b^{2}+c^{2}\right) \left( \dfrac {1} {a}+\dfrac {1} {b}+\dfrac {1} {c}\right)\geq a+b+c+\sqrt {a^{2}+b^{2}+c^{2}}$$

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## 1 Answer

Because the inequality is homogeneous, we can assume that $a^2 + b^2 + c^2 = 1$. Then by the power mean inequality $$\frac{a+b+c}{3} \le \sqrt{\frac{a^2 + b^2 + c^2}{3}} = \frac{1}{\sqrt{3}}$$ and $$\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \le \sqrt{\frac{a^2 + b^2 + c^2}{3}} = \frac{1}{\sqrt{3}}.$$ Thus $$a+b+c+1 \le 1 + \sqrt{3}= \frac{1 + \sqrt{3}}{3\sqrt{3}} 3\sqrt{3} \le \frac{1 + \sqrt{3}}{3\sqrt{3}} \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right).$$

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