An inequality in positive real continuous function I proposed my conjecture as follows:
Let $f(x)$ is a positive real continuous function that is convex on $[m, M]$, let $m \le x_i \le M$, for $i=1,2,...,n$ then show that
$$\frac{f(x_1)+f(x_2)+.....+f(x_n)}{n} \le \frac{f(M)+ f(m)}{2f\left(\frac{M+m}{2}\right)} f\left(\frac{x_1+x_2+....x_n}{n}\right)$$
Equality holds if only if $m=x_1=x_2=....=x_n=M$
 A: What if you fix $\epsilon>0$ and consider $f:[0,3]\rightarrow\mathbb{R}$ defined by $f(x) = (x-1)^2 + \epsilon$?  So $m=0, M=3$.  Let $x_1=0, x_2=2$.  So: 
\begin{align}
\frac{f(x_1)+f(x_2)}{2}  &= \frac{f(0)+f(2)}{2} = 1 + \epsilon\\
f(\frac{x_1+x_2}{2}) &= f(1) = \epsilon \\
\frac{f(M)+f(m)}{2} &= \frac{f(3)+f(0)}{2} = (5/2)+\epsilon\\
f(\frac{M+m}{2}) &= f(1.5) = (1/4)+\epsilon
\end{align}
So for a counter-example to your conjecture, we just find an $\epsilon>0$ such that: 
$$ 1+\epsilon > \left(\frac{(5/2)+\epsilon}{(1/4)+\epsilon}\right)\epsilon $$
which is true for all sufficiently small $\epsilon>0$. 
A: You made a second conjecture in the comment to my first answer, so I am answering that here.  The conjecture is: 
Conjecture:
$$ \frac{f(x_1)+...+f(x_n)}{n} - f(\frac{x_1+...+x_n}{n}) \leq \frac{f(M)+f(m)}{2}- f(\frac{M+m}{2}) $$
whenever $f$ is continuous and convex over the interval $[m,M]$ and $m \leq x_i\leq M$ for all $i$. 
Counter-example:
Consider $m=0,M=1$.  Consider $f:[0,1]\rightarrow\mathbb{R}$ defined with 2 piecewise linear segments over the intervals $[0,3/4]$ and $[3/4,1]$ with: 
\begin{align}
f(0) &= 2\\
f(3/4) &= 1\\
f(1) &= 2
\end{align}
Specifically: 
$$ f(x) =  \left\{ \begin{array}{ll}
-(4/3)x + 2 &\mbox{ if $x \in [0, 3/4]$} \\
4x-2  & \mbox{ if $x \in [3/4,1]$} 
\end{array}
\right. $$
Then: 
$$ \frac{f(M)+f(m)}{2} - f(\frac{M+m}{2})= \frac{f(1)+f(0)}{2} - f(1/2) = 2/3$$
Now let $x_1=0, x_2=x_3=x_4=1$.  So:
$$ \frac{f(x_1)+...+f(x_4)}{4} - f(\frac{x_1+...+x_4}{4}) = 2-f(3/4)=1$$
