Find maximum of function $A=\sum _{cyc}\frac{1}{a^2+2}$ Let $a,b,c\in R^+$ such that $ab+bc+ca=1$. Find the maximum value of $$A=\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}$$

I will prove $A\le \dfrac{9}{7}$ and the equality occurs when $a=b=c=\dfrac{1}{\sqrt3 }$
$$\frac{\sum _{cyc}\left(b^2+2\right)\left(c^2+2\right)}{\Pi _{cyc}\left(a^2+2\right)}\le \frac{9}{7}$$
$$\Leftrightarrow 7\sum _{cyc}a^2b^2+28\sum _{cyc}a^2+84\le 9a^2b^2c^2+18\sum _{cyc}a^2b^2+36\sum _{cyc}a^2+72 (1)$$
Let $a+b+c=3u;ab+bc+ca=3v^2=1;abc=w^3$ then we need to prove
$$9w^6+11\left(9v^4-6uw^3\right)+8\left(9u^2-6v^2\right)-12\ge 0$$
$$\Leftrightarrow 9w^6+11\left(27v^6-18uv^2w^3\right)+8\left(81u^2v^4-54v^6\right)-324v^6\ge 0$$
$$\Leftrightarrow 9w^6-198uv^2w^3-459v^6+648u^2v^4\ge 0$$
We have: $$f'\left(w^3\right)=18\left(w^3-11uv^2\right)\le 0$$
So $f$ is a decreasing function of $w^3$ it's enough to prove it for an equality case of two variables. Assume $a=b\rightarrow c=\dfrac{1-a^2}{2a}$ 
$$(1)\Leftrightarrow \frac{(a^2+2)(a^2+4)(3a^2-1)^2}{4a^2}\ge0$$
Please check my solution. It's the first time i use $u,v,w$, if i have some mistakes, pls fix for me. Thanks!
 A: Your solution is right, but if you  want to get all points for this problem, you need to explain, why it's enough to prove the starting inequality for equality case of two variables. It's not enough to say that it's true by uvw.
After words: $f$ decreases you can write the following.
Thus, it's enough to prove our inequality for a maximal value of $w^3$.
Now, $a$, $b$ and $c$ are  roots of the equation
$$(x-a)(x-b)(x-c)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$g(x)=w^3,$$ where
$$g(x)=x^3-3ux^2+3v^2x.$$
We see that
$$g'(x)=3x^2-6ux+3v^2=3\left(x-(u-\sqrt{u^2-v^2})\right)\left(x-(u+\sqrt{u^2-v^2})\right),$$
which says that $g$ gets a local maximum for $x=u-\sqrt{u^2-v^2}$ and the equation
$$g(x)=w^3$$ has three positive roots for $$\max\left\{0,g\left(u+\sqrt{u^2+v^2}\right)\right\}<w^3\leq g\left(u-\sqrt{u^2-v^2}\right)$$ and $w^3$ gets a maximal value, when the graph of $y=w^3$ is a tangent line to the graph of $g$, which happens for equality cases of two variables.
We got even that these variables should be equal to $u-\sqrt{u^2-v^2}.$
The rest you wrote already.
A: $$a=\tan \left(\frac {\alpha}{2}\right), b=\tan \left(\frac {\beta}{2}\right),    c=\tan \left(\frac {\gamma}{2}\right)\ \ \ \ (\alpha+\beta+\gamma= \pi)$$
$$A=\dfrac { \cos^2 \left(\frac {\alpha}{2}\right) }{1+  \cos^2 \left(\frac {\alpha}{2}\right)  }+ \dfrac { \cos^2 \left(\frac {\beta}{2}\right) }{1+  \cos^2 \left(\frac {\beta}{2}\right)  }+ \dfrac { \cos^2 \left(\frac {\gamma}{2}\right) }{1+  \cos^2 \left(\frac {\gamma}{2}\right)  } \le  \dfrac {3t}{3+t}\le \dfrac {9}{7} $$
$$t= \cos^2 \left(\frac {\alpha}{2}\right) + \cos^2 \left(\frac {\beta}{2}\right) +  \cos^2 \left(\frac {\gamma}{2}\right)=\dfrac {3}{2}  + \dfrac {1}{2} (\cos (\alpha)+\cos (\beta)+\cos (\gamma)) =\dfrac {3}{2}+\dfrac {R+r}{2R} \le \dfrac {9}{4}$$
