Mapping of Inverse functions. 
Let $\mathsf{f(x)}$ be a differential and an invertible function, such that $\mathsf{f''(x)>0}$ and $\mathsf{f'(x)>0}$.
Prove that$$\mathsf{ f^{-1}\left(\frac{x_1 + x_2 +x_3}{3}  \right) > \frac{f^{-1}(x_1)+f^{-1}(x_2)+f^{-1}(x_3)}{3}}$$

I have no clue, how to start it. I think a graphical solution can be obtained but I am confused about the graph of the inverse function.

I guess $f$ is defined on $\mathbb R$ and $x_1,x_2,x_3$ are not equals. Since $f''(x)>0$, $f$ is stricly convex, hence 
$$f\left(\frac{f^{-1}(x_1)}3+\frac{f^{-1}(x_3)}3+\frac{f^{-1}(x_3)}3\right)< \frac{x_1+x_2+x_3}3,$$
and since $f$ is strictly increasing, so is $f^{-1}$. It gives 
$$\frac{f^{-1}(x_1)}3+\frac{f^{-1}(x_3)}3+\frac{f^{-1}(x_3)}3<f^{-1}\left(\frac{x_1+x_2+x_3}3\right).$$
