Let $a,b,c>0$ such that $a+b+c=3$, find max/min of $\frac1{a^2+1}+\frac1{b^2+1}+\frac1{c^2+1}$ 
Let $a,b,c>0$ be real numbers such that $a+b+c=3$, find the maximum and minimum of:
  $$\frac1{a^2+1}+\frac1{b^2+1}+\frac1{c^2+1}$$

I've tried using AM-GM-HM, but I can't find the maximum or the minimum.
Using intuition, substituting $a=b=c=1$ should give us one of the optimums, but I cannot prove that it is the optimum.
 A: Let $f(x) = \frac{1}{1+x^2}$ and $\mu = \frac{1}{\sqrt{3}}$. 
WOLOG, we will assume $a \le b \le c$.
Notice for any $x \ge 0$,
$$f(x) + \frac12(x-2) = \frac{x(x-1)^2}{2(x^2+1)} \ge 0$$
Substitute $x$ by $a, b, c$ and sum, we obtain
$$S(a,b,c) \stackrel{def}{=} f(a)+f(b)+f(c) \ge \frac12(6-(a+b+c)) = \frac32$$
Since this lower bound $\frac32$ is achieved at $a = b = c = 1$, this is the minimum we seek.
For maximum, we use the fact
$$f''(x) = 2\frac{(3x^2-1)}{(x^2+1)^3}
\quad\text{ is }\quad
\begin{cases}
< 0, & x < \mu\\
= 0, & x = \mu\\
> 0, & x > \mu
\end{cases}$$
to conclude $f(x)$ is strictly concave over $[0,\mu)$ and strictly convex over $(\mu,\infty)$. 
As a result of this, when $(a,b,c)$ is a configuration which maximizes $S(a,b,c)$, we have


*

*At most one of $a,b,c$ lies inside $(\mu,\infty)$.
Assume the contrary, let's say $\mu < b \le c$. 


*

*When $b < c$, 
$f''(x) > 0$ over $(\mu,\infty)$ implies $f'(x)$ is increasing there.
This means $f'(b) < f'(c)$. For small enough $\epsilon > 0$, we have
$$S(a, b-\epsilon,c+\epsilon) = S(a,b,c) + (f'(c)-f'(b))\epsilon + O(\epsilon^2) > S(a,b,c)$$
This contradicts with our choice of $a,b,c$ to maximize $S(a,b,c)$.

*When $b = c$, $f(x)$ is strictly convex over $(\mu,\infty)$ and Jensen's inequality tell us for small enough $\epsilon > 0$,
$$S(a, b-\epsilon, b+\epsilon) > S(a,b,b)$$ 
This again contradicts with our choice of $a, b, c$.


*For those $a,b,c$ lies inside $[0,\mu]$, we can assume they share a common value.
This is because $f(x)$ is concave over $[0,\mu]$, Jensen's inequality
tell us if we replace those $a,b,c$ inside $[0,\mu]$ by their average, it will only increase $S(a,b,c)$.
This leaves us with two possible configurations for $(a,b,c)$. Either


*

*$a = b = c \le \mu$

*$a = b \le \mu < c$


It is easy to see the first scenario gives us $a = b = c = 1$. This gives us the minimum instead of maximum. For the second scenario, we can look at the  parametrization $(a,b,c) = (t,t,3-2t)$.
We find
$$S(a,b,c) = S(t,t,3-2t) = \frac{2}{1+t^2} + \frac{1}{1+(3-2t)^2}$$
Differentiate RHS gives us
$$\frac{4(3-2t)}{1+(3-2t)^2)^2}-\frac{4t}{(1+t^2)^2}
= -\frac{3(t-1)(6t^4-27t^3+49t^2-33t+1)}{((t^2+1)^2 (2t^2-6t+5)^2}$$
A plot of it shows that it is mostly negative over $[0,1)$.
It is positive over $[0,\lambda)$ where $\lambda
\approx 0.031776261136412972$ is the smallest positive real root of the quartic polynomial: 
$$6t^4-27t^3+49t^2-33t+1$$
$S(t,t,3-2t)$ is increasing on $[0,\lambda]$ and decreasing on $[\lambda,\mu]$. 
The maximum of $S(a,b,c)$ is achieved at 
$$(\lambda,\lambda,3-2\lambda) \approx (0.031776261136412972,0.031776261136412972,2.936447477727174)$$
with value $\approx 2.101903255548146$.
