If $f(x)$ is monotonically decreasing function and $f^{''}(x)>0$. Assuming $f^{-1}(x)$ exists prove that, $$\dfrac{f^{-1}(x_1)+f^{-1}(x_2)}{2}>f^{-1}\bigg(\dfrac{x_1+x_2}{2}\bigg)$$
My attempts:
As $f(x)$ decreases hence when $x\uparrow\implies y\downarrow$. Also $f^{-1}(x)$ is a reflection of $f(x)$ in $y=x\implies$ to find $f^{-1}$ we interchange $x\rightarrow y$. So , for $f^{-1}$ as $x\downarrow\implies y\uparrow$. So both should have same type of curvature. I've considered $f^{-1}(x)$ as follows:
From above graph, $E$ is midpoint of $C$ and $D$, $EF\perp x-\text{axis}$ hence $x_E=x_F$, from using curvature of $f^{-1}(x)$ we can see that the ordinate of $F$ is less than that of $E$ $$\implies\dfrac{f^{-1}(x_1)+f^{-1}(x_2)}{2}>f^{-1}\bigg(\dfrac{x_1+x_2}{2}\bigg)$$
Please point out all flaw statements I assumed above and want to see other elegant methods to solve the same problem. Thanks!
$f$ is strictly convex, so, for $x_1\ne x_2$, $$f\left( \frac{f^{-1}(x_1)+f^{-1}(x_2)}{2}\right)< \frac{f\circ f^{-1}(x_1)+f\circ f^{-1}(x_2)}{2}=\frac{x_1+x_2}{2}$$ Now, $f^{-1}$ is monotone deceasing so $$f^{-1}\circ f\left( \frac{f^{-1}(x_1)+f^{-1}(x_2)}{2}\right)> f^{-1}\left(\frac{x_1+x_2}{2}\right)$$ That is $$\frac{f^{-1}(x_1)+f^{-1}(x_2)}{2}> f^{-1}\left(\frac{x_1+x_2}{2}\right).$$ So, we have: $f$ convex and decreasing implies $f^{-1}$ convex and decreasing. Generally, $f$ convex does not imply $f^{-1}$ convex, take for example $f:[0,1]\to\mathbb{R},f(x)=x^2$.