$f(x)$ is monotonically increasing function for all real $x$, $f’’(x)>0$ and $f^{-1}$ exists, then prove that $\frac{f^{-1}(x_1)+f^{-1}(x_2)}{2}<f^{-1}(\frac{x_1+x_2}{2})$
The conclusions that can be drawn from the given data is that $f(x)$ is monotonous and increasing, so $f^{-1}$ is also monotonous and increasing. Also $f(x)$ is concaved upwards, so $f^{-1} (x)$ is concaved downwards. How do I solve from here?
Visualisation helps.$g(x)=f^{-1}(x)$.
Imgine $g(x)$ concave downward.Let P be the point $(x_1,g(x_1))$ and Q be $(x_2,g(x_2))$.
Thus the point $(\frac{x_1+x_2}{2},\frac{g(x_1)+g(x_2)}{2})$ is the mid point of of segment $PQ$.
Also a perpendicular line line drawn from $(\frac{x_1+x_2}{2},0)$ meets the curve in $(\frac{x_1+x_2}{2},g(\frac{x_1+x_2}{2}))$.
It is easy to see that this point lies above midpoint of $PQ$