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!

$$a=\tan \left(\frac {\alpha}{2}\right), b=\tan \left(\frac {\beta}{2}\right),    c=\tan \left(\frac {\gamma}{2}\right)\ \ \ \ (\alpha+\beta+\gamma= \pi)$$
$$A=\dfrac { \cos^2 \left(\frac {\alpha}{2}\right) }{1+  \cos^2 \left(\frac {\alpha}{2}\right)  }+ \dfrac { \cos^2 \left(\frac {\beta}{2}\right) }{1+  \cos^2 \left(\frac {\beta}{2}\right)  }+ \dfrac { \cos^2 \left(\frac {\gamma}{2}\right) }{1+  \cos^2 \left(\frac {\gamma}{2}\right)  } \le  \dfrac {3t}{3+t}\le \dfrac {9}{7} $$
$$t= \cos^2 \left(\frac {\alpha}{2}\right) + \cos^2 \left(\frac {\beta}{2}\right) +  \cos^2 \left(\frac {\gamma}{2}\right)=\dfrac {3}{2}  + \dfrac {1}{2} (\cos (\alpha)+\cos (\beta)+\cos (\gamma)) =\dfrac {3}{2}+\dfrac {R+r}{2R} \le \dfrac {9}{4}$$
