<s>[INST] Proof about an inequality We have $a,b,c>0$, prove that:
$$\dfrac {1+\sqrt {3}} {3\sqrt {3}}\left( a^{2}+b^{2}+c^{2}\right) \left( \dfrac {1} {a}+\dfrac {1} {b}+\dfrac {1} {c}\right)\geq a+b+c+\sqrt {a^{2}+b^{2}+c^{2}}$$
 [/INST] 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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