Find maximize of the function $\frac{a}{1+a^2}+\frac{b}{1+b^2}-\frac{1}{c^2+1}$

Question

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

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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 $P$?

Answer 1

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 $a=b$: $$P(a)=\frac{2a}{a^2+1}-\frac{4a^2}{(a^2+1)^2}\stackrel{x=\frac{2a}{a^2+1}}{=}x-x^2\le \frac14 \Rightarrow x=\frac12=\frac{2a}{a^2+1} \Rightarrow a=2-\sqrt{3}.$$

Answer 2

Let $a=\tan\frac{\alpha}{2}$, $b=\tan\frac{\beta}{2}$ and $c=\tan\frac{\gamma}{2},$ where $\left\{\frac{\alpha}{2},\frac{\beta}{2},\frac{\gamma}{2}\right\}\subset(0^{\circ},90^{\circ}).$

Thus, $\alpha+\beta+\gamma=180^{\circ}$ and $$P=\frac{1}{2}\sin\alpha+\frac{1}{2}\sin\beta-\cos^2\frac{\gamma}{2}=\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}-\cos^2\frac{\gamma}{2}=$$ $$=\cos\frac{\gamma}{2}\cos\frac{\alpha-\beta}{2}-\cos^2\frac{\gamma}{2}\leq\cos\frac{\gamma}{2}-\cos^2\frac{\gamma}{2}$$ Can you end it now?

I got that the maximal value it's $\frac{1}{4}.$