What's the maximum value of $\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}$ given that: $abc+a+c=b$ 
Given that: $abc+a+c=b$. What's the maximum value of $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}.$$

(DO NOT use trigonometric methods)
 A: prove that $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}\le \frac{10}{3}$$ and the equal sign holds if $$a=\frac{2}{3}\left(\frac{1}{2\sqrt{2}}-\sqrt{2}\right),b=-\sqrt{2},c=-\frac{1}{2\sqrt{2}}$$
A: For  $a=\frac{1}{\sqrt2}$, $b=\sqrt2$ and $c=\frac{1}{2\sqrt2}$ we'll get a value $\frac{10}{3}$.
We'll prove that it's a maximal value.
Indeed, the condition gives $c(ab+1)=b-a$.
If $ab=-1$ then $a=b$ and $a^2+1=0$, which is impossible.
Thus, $ab\neq-1$, $c=\frac{b-a}{ab+1}$ and we need to prove that
$$\frac{2}{1+a^2}-\frac{2}{1+b^2}+\frac{3}{1+\left(\frac{b-a}{1+ab}\right)^2}\leq\frac{10}{3}$$ or
$$\frac{2(b^2-a^2)}{(1+a^2)(1+b^2)}+\frac{3(1+ab)^2}{(1+a^2)(1+b^2)}\leq\frac{10}{3}$$ or
$$(a^2+4)b^2-18ab+16a^2+1\geq0,$$
for which it's enough to prove that
$$81a^2-(a^2+4)(16a^2+1)\leq0,$$
which is
$$(2a^2-1)^2\geq0.$$
Done!
I got a value $\frac{10}{3}$ and values $a$, $b$ and $c$ by the following way.
Let $k$ be a maximal value of our expression.
Thus, the following inequality is true for all values of $a$ and $b$.
$$\frac{2(b^2-a^2)}{(1+a^2)(1+b^2)}+\frac{3(1+ab)^2}{(1+a^2)(1+b^2)}\leq k$$ or
$$(ka^2-3a^2+k-2)b^2-6ab+ka^2+2a^2+k-3\geq0.$$
Thus, $k\geq3$ and since we have a quadratic inequality of $b$, we need that the equality
$$9a^2-(ka^2-3a^2+k-2)(ka^2+2a^2+k-3)\leq0$$ or
$$(k^2-k-6)a^4+2(k^2-3k-2)a^2+k^2-5k+6\geq0$$
will be true for all value of $a$.
But the last inequality is a quadratic inequality of $a^2$, which says that we need
$$(k^2-3k-2)^2-(k^2-k-6)(k^2-5k+6)\leq0$$ or
$$3k-10\geq0,$$
which gives a value $\frac{10}{3}$ and from here we can get the case of the equality occurring. 
