How prove $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$

An analytical proof is proposed below :

By AM-GM, $1=\frac{3}{a^2+2}+\frac{3}{b^2+2}+\frac{3}{c^2+2} \geq \frac{9}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}$

By AM-GM, $7 = \frac{3a^2}{a^2+2}+\frac{3b^2}{b^2+2}+\frac{3c^2}{c^2+2} \geq \frac{9\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}$

As everything is positive, we can safely divide the inequalities with each other.

$\frac{7}{1} \geq \frac{\frac{9\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}}{\frac{9}{\sqrt[3]{(a^2+2)(b^2+2)(c^2+2)}}}$

$\Rightarrow 7 \geq (abc)^{\frac{2}{3}}$

$\Rightarrow \frac{1}{\sqrt[3]{abc}} \geq \frac{1}{\sqrt{7}}$

By AM-GM, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{3}{\sqrt[3]{abc}}\geq \frac{3}{\sqrt{7}}$

$\therefore \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{3}{\sqrt{7}}$