Finding maximum of $A=\frac a{2+bc}+\frac b{2+ca}+\frac c{2+ab}$ 
Let $a,b,c\ge 0$ satisfy $a^2+b^2+c^2=2$. Find maximum of $$A=\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}.$$

I see $\max A=1$ and it occurs when $(a,b,c)=(1,1,0)$ and its permutation. So I will prove this inequality:$$\frac{a}{2+bc}\le \frac{a}{a+b+c} \quad \text{or} \quad 2+bc\ge a+b+c.$$
It is true because
$$2(2+2bc)=(1+1)(a^2+(b+c)^2)\ge (a+b+c)^2.$$
Is it right? And I want a new method.

It is true that $2(2+2bc)=(1+1)(a^2+(b+c)^2)\ge (a+b+c)^2$, but it doesn't prove your inequality because $2 \le a+b+c$ is not proved. Actually, it is not true. Try $a=b=c=\sqrt{\frac{2}{3}}$.
Instead, $2+bc\ge a+b+c$ is equivalent to$$2-b-c+bc\ge a$$or$$(2-b-c+bc)^2\ge a^2$$or$$b^2 c^2 - 2 b^2 c + 2b^2 - 2 b c^2 + 6 bc - 4 b + 2c^2 - 4 c +2
\ge0$$or$$(b-1)^2(c-1)^2+(b+c-1)^2\ge0$$which is obvious.
