Find maximize of the function $\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$ Let $a,b\in R^+$ such that $ab+bc+ca=1$. Find the maximize of $$P=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$$

By Wolframalpha i can see that if $a=b=2-\sqrt 3;c=\sqrt 3$ we will have $P=\dfrac 1 4$
WLOG $a\le b$. I proved that $$P=f(a,b,c)\le f(a,a,c)\le 0$$
$$\Leftrightarrow -(a^2-4a+1)^2\le 0$$
My proof is based on the value of $P$ so it's inconvenient if i dont know value of $P$. This inequality is not homogeneous and symmetric so i dont any idead to solve it
Can you help me solve it without using the equality and value of $P$? 

Find $c$:
$$ab+bc+ca=1 \Rightarrow c=\frac{1-ab}{a+b}$$
Sub to $P$:
$$\begin{align}P&=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}=\\
&=\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{(a+b)^2}{(a^2+1)(b^2+1)}=\\
&=\frac{(a+b)(a-1)(b-1)}{(a^2+1)(b^2+1)}\end{align}$$
Now it is symmetric and the maximum is achieved at $a=b$:
$$P(a)=\frac{2a}{a^2+1}-\frac{4a^2}{(a^2+1)^2}\stackrel{x=\frac{2a}{a^2+1}}{=}x-x^2\le \frac14 \Rightarrow x=\frac12=\frac{2a}{a^2+1} \Rightarrow a=2-\sqrt{3}.$$
