Proving $f^{-1}$ is concave. Let $f:(a,b)\to \mathbb R$ be an invertible, non-decreasing convex function. Prove $f^{-1}$ is concave. There are all these theorems and rules but I can't make it there... Would appreciate your help.
 A: Graph of $f$ and $f^{-1}$ are symmetric about $y=x$, so if $f$ is non decreasing convex fnction then $f^{-1}$ must be concave.
A: Let $x,x'\in(a,b)$ and $y=f(x),y'=f(x')$ and $\lambda\in[0,1]$ then since $f$ is non decreasing then $f^{-1}$ is non-decreasing so the desired result is based on the definition of the convexity:
$$f(\lambda x+(1-\lambda)x')\le \lambda f(x)+(1-\lambda)f(x')\iff \lambda x+(1-\lambda)x'\le f^{-1}(\lambda f(x)+(1-\lambda)f(y)\\\iff \lambda f^{-1}(y)+(1-\lambda)f^{-1}(y')\le f^{-1}(\lambda y+(1-\lambda)y')$$
A: If $f$ is non-decreasing and invertible, so is $f^{-1}$.
Since $f$ is convex, we have, for $\lambda \in [0,1]$,
$f(\lambda x + (1-\lambda )y) \le \lambda f(x) + (1-\lambda) f(y)$. Applying
$f^{-1}$ gives
$\lambda x + (1-\lambda )y) \le f^{-1}(\lambda f(x) + (1-\lambda) f(y))$.
Now suppose $x=f^{-1}(s), y=f^{-1}(t)$, then this gives
the desired result.
Alternative: Here is a more geometric approach (inspired by Neeraj's answer):
Again, this hinges on the fact that if $f$ is non-decreasing, then so
is $f^{-1}$.
Note that $(x,y) \in \operatorname{epi} f$  iff $y \ge f(x)$ iff
$f^{-1}(y) \ge x$ iff $-x \ge -f^{-1}(y)$ iff $(y,-x) \in \operatorname{epi} (-f^{-1})$.
Since $f$ is convex and the transformation $\phi(x,y) = (y,-x)$ is linear, we see that $\operatorname{epi} (-f^{-1}) = \phi (\operatorname{epi} f)$ and so
it is convex and hence $-f^{-1}$ is convex. Taking account of the minus sign, we see that $f^{-1}$ is concave.
A: Here is a proof that is probably not the most useful for the OP and requires extra assumptions.  But it's a neat chain rule application.
Suppose additionally that $f$ is twice-differentiable.  Since the graph of $f$ is convex, we have $f''(x) > 0$ for all $x$ within in the domain of $f$.  Since $f$ is non-decreasing, $f'(x) > 0$ for all $x$ too.
(I think the convexity and monotonicity assumptions together rule out $f'(x) = 0$. The canonical counterexample $f(x) = x^3$ doesn't satisfy the concavity assumption.  In case I am wrong, we can go ahead and assume $f'(x) > 0$ for all $x$.)
The inverse function theorem tells us that for all $y$ in the range of $f$,
$$
(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}
$$
Take another derivative, and apply the chain rule:
\begin{align*}
    (f^{-1})''(y) &= \frac{-1}{(f'(f^{-1}(y)))^2} \cdot f''(f^{-1}(y)) \cdot (f^{-1})'(y) \\
&= \frac{-1}{(f'(f^{-1}(y)))^2} \cdot f''(f^{-1}(y)) \cdot \frac{1}{f'(f^{-1}(y))} \\
&= \frac{- f''(f^{-1}(y))}{(f'(f^{-1}(y)))^3}
\end{align*}
Since $f'$ and $f''$ are positive, we have $(f^{-1})''(y) < 0$.  So the graph of $f^{-1}$ is concave.
