How prove $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$ Let $a,b,c&gt;0$,and such
  $$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}$$
  show that
  $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}$$


my try: since
$$\dfrac{1}{2+a^2}=\dfrac{1}{2}\left(1-\dfrac{a^2}{a^2+2}\right)$$
so
$$\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}=\dfrac{7}{3}$$
we only prove 

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}$

and I want use Cauchy-Schwarz inequality to prove it,But I can't works,such
$$\left(\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}\right)(a^2+2+b^2+2+c^2+2)\ge (a+b+c)^2$$
$$(a^2+b^2+c^2+6)\ge \dfrac{3}{7}(a^2+b^2+c^2+2ab+2bc+2ac)$$
$$\Longrightarrow 4(a^2+b^2+c^2)+42\ge 6(ab+bc+ac)$$
and let $$p=a+b+c,q=ab+bc+ac,r=abc$$
so
$$2p^2+21\ge 7q$$
and we only prove
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ac}{abc}=\dfrac{q}{r}\le\dfrac{3}{\sqrt{7}}$$
maybe this is not true.

But this not usefull to solve this problem .

Thank you
An analytical proof is proposed below :
This might not be the "proper" way, but it can be seen that both the expressions are symmetrical in all terms. And we need the maximum of the second eqn. This can happen only when all the terms contribute equally to the sum, if it were not so then, let 1/a&lt;1/b&lt;1/c, then we can make b and c to a, and get higher sum. So, putting it back in the first eqn. we get $ \frac{3}{a^2+2} = 1/3$, solving which we get $ a = \sqrt{7}$ or $-\sqrt{7} $. Choosing the positive root, as we need maximum values, we get $ \frac{1}{a} + \frac{1}{b}+\frac{1}{c} = \frac{3}{\sqrt{7}}$

Missed JJacquelin's answer, by a few minutes. It is more analytical, than mine.