# 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.

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Hint: Did you learn something about the relationship between the sign of the second derivative of a function and convexity? If yes, can you calculate the second derivative of $f^{-1}$ in terms of $f, f^'$ and $f{''}$ ? Can you determine the sign of it? – user20266 Jan 2 '12 at 21:09

## 2 Answers

This should help : 'Inverse of Convex Strictly Monotone Function'.
Use the definition of concavity with $\alpha=1/3, \beta=2/3$ and so on...

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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 $$\frac{f^{-1}(x_1)}3+\frac{f^{-1}(x_3)}3+\frac{f^{-1}(x_3)}3<f^{-1}\left(\frac{x_1+x_2+x_3}3\right).$$

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