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

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

Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>\frac{3}{\sqrt7}$, $\frac{1}{a}=\frac{kx}{\sqrt7}$, $\frac{1}{b}=\frac{ky}{\sqrt7}$ and $\frac{1}{c}=\frac{kz}{\sqrt7}$, where $k>0$ and $x+y+z=3$.

Hence, $k>1$ and $\frac{1}{3}=\sum\limits_{cyc}\frac{1}{a^2+2}=\sum\limits_{cyc}\frac{1}{\frac{7}{k^2x^2}+2}>\sum\limits_{cyc}\frac{1}{\frac{7}{x^2}+2}=\sum\limits_{cyc}\frac{x^2}{7+2x^2}$,

which is a contradiction because we'll prove now that $\sum\limits_{cyc}\frac{x^2}{7+2x^2}\geq\frac{1}{3}$.

Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $$4x^2y^2x^2+\sum\limits_{cyc}(8x^2y^2+7x^2)-49\geq0$$ or $$4w^6+8(9v^4-6uw^3)u^2+7(9u^2-6v^2)u^4-49u^6\geq0$$ or $f(w^3)\geq0$, where $f(w^3)=2w^6-24u^3w^3+7u^6-21u^4v^2+36u^2v^4$.

We see that $f'(w^3)=4w^3-24u^3<0$, which says that it's enough to prove that

$f(w^3)\geq0$ for a maximal value of $w^3$, which happens for equality case of two variables.

Let $y=x$ and $z=3-2x$.

Hence, $\sum\limits_{cyc}\frac{x^2}{7+2x^2}\geq\frac{1}{3}\Leftrightarrow(x-1)^2(2x-1)^2\geq0$.

Done!

Also we can use the Vasc's LCF Theorem.