How find this maximum $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{1+c^2}$ let $a,b,c>0$ ,and such $a+b+c=3$.prove that
$$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{1+c^2}<\dfrac{11}{5}$$
if this problem  find this minimum,then 
$$f(x)=\dfrac{1}{x^2+1}\ge ax+b$$
where $a=f'(1),a+b=f(1)$
since $$f'(x)=-\dfrac{2x}{(1+x^2)^2}\Longrightarrow f'(1)=-\dfrac{1}{2}$$
so
$$a=-\dfrac{1}{2},b=1$$
and
$$\dfrac{1}{1+x^2}+\dfrac{1}{2}x-1=\dfrac{x(x-1)^2}{(x^2+1)^2}\ge 0$$
then
$$\dfrac{1}{1+x^2}\ge-\dfrac{1}{2}x+1$$
so
$$\sum_{cyc}\dfrac{1}{1+a^2}\ge\sum_{cyc}-\dfrac{1}{2}a+1=-\dfrac{1}{2}(a+b+c)+3=\dfrac{3}{2}$$
But for maximum,I can't prove it.Thank you
 A: First show that $\dfrac{1}{1+x^2} < \dfrac{63 - 19x}{60}$ for $0 < x < 3$.
Then $\text{LHS} < \dfrac{3\cdot63 - 19(a + b + c)}{60} = \dfrac{189 - 57}{60} = \text{RHS}$.
A: By symmetry, it just suffices to look at the case $0 \le a \le b \le c$ where $a + b + c = 3$.
Let $f(x) = \frac{1}{1+x^2}$ and $F(a,b,c) = f(a)+f(b)+f(c)$. Notice
$$\frac{d^2}{dx^2} f(x) = \frac{2(3x^2-1)}{(1+x^2)^3} \longrightarrow
\begin{cases} > 0,&\text{ for } x > \frac{1}{\sqrt{3}}\\< 0,&\text{ for } x <\frac{1}{\sqrt{3}}\end{cases}$$
The function $f(x)$ is concave on $[0,\frac{1}{\sqrt{3}}]$ and convex on $[\frac{1}{\sqrt{3}},3]$.
Since $a + b + c = 3$, we know $c \ge 1$. This implies $c \in (\frac{1}{\sqrt{3}}, 3]$ where $f(x)$ is convex. If $b$ also belongs to this interval, then we can  increase the value of $F(a,b,c)$ by decreasing $b$ and increasing $c$ symmetrically w.r.t their mean $\frac{3-a}{2}$ until either $b$ hits $\frac{1}{\sqrt{3}}$ or $c$ hits $3$. It is easy to see $b$ will hit $\frac{1}{\sqrt{3}}$ first. From this, we can conclude when $b > \frac{1}{\sqrt{3}}$,
$$F(a,b,c) \le F(a, \frac{1}{\sqrt{3}}, 3-a-\frac{1}{\sqrt{3}})$$
If $a > \frac{1}{\sqrt{3}}$, the new $b$ will be smaller than $a$. We can flip the roles of $a$ and $b$ and repeat above trick. At the end, we will find a pair of $a', b'
\in [0, \frac{1}{\sqrt{3}}]$ such that
$$F(a,b,c) \le F(a',b',c')\quad\text{ where }\;\;c' = 3 - a'-b'$$
Since $f(x)$ is concave there, we have
$$F(a',b',c') \le F(\frac{a'+b'}{2},\frac{a'+b'}{2},c')$$
This means in order to find the optimal $(a,b,c)$ which maximize $F(a,b,c)$, one only need to look at those points of the form
$$(a,b,c) = (x,x,3-2x)\quad\text{ where } x \in [0,\frac{1}{\sqrt{3}}]$$
For such points, $F(a,b,c)$ reduces to
$$F(x,x,3-2x) = \frac{2}{1+x^2} + \frac{1}{1 + (3-2x)^2}$$
Let us call this function as $G(x)$. Notice
$$\frac{dG(x)}{dx} = \frac{4(3-2x)}{(1+(3-2x)^2)^2} - \frac{4x}{(1+x^2)^2}$$
We have
$$G'(0) = \frac{3}{25}> 0\quad\text{ and }\quad G'(\frac{1}{\sqrt{3}}) = \frac{1944\sqrt{3}-3645}{1588\sqrt{3}-2448} \sim -0.91866538126469 < 0$$
So the optimal $(a,b,c) = (x,x,3-2x) $ occurs for a $x \in (0,\frac{1}{\sqrt{3}})$. 
We can determine the corresponding value of $x$ by solving
$$G'(x) = 0\quad\iff\quad 6x^4-27x^3+49x^2-33x+1 = 0$$
The polynomial in R.H.S has 2 real roots, only one of it $\sim 0.031776261136413$ falls inside the interval $[0,\frac{1}{\sqrt{3}}]$. So the maximum value of $F(a,b,c)$ is about
$$\max F(a,b,c) \sim G(0.031776261136413) \sim 2.101903255548146 < \frac{11}{5}$$
