If $f:[0,1] \rightarrow [0,1]$ such that $f'(x)≥0$, $f''(x)≤0$ then prove that $$f(x)\cdot f^{-1}(x)-x^2≤0$$
I was able to form a graphical solution but it doesn't feel rigorous enough to me, can someone else come up with some alternate proof?
My ad-hoc method:
Take $A (t,f(t))$, $B(t,f^{-1}(t))$ and $B' (f^{-1}(t),t)$ (note $B'$ lies on $f(x)$). Now the graph of $f(x)$ will look something like $\sqrt{x}$ and $f^{-1}$ will be it's reflection about $y=x$ so once we make the proper diagram we will see that
slope $OB'≥$ slope $OA$ $$\implies \frac{x}{f^{-1}(x)}≥\frac{f(x)}{x}$$ $$\implies f(x)\cdot f^{-1}(x)-x^2≤0$$
We first note that the claim follows if we can show that $\frac{f(x_1)}{x_1} \geq \frac{f(x_2)}{x_2}$ whenever $x_1 \leq x_2$. To see this first observe that it implies $\frac{f(x)}{x} \geq \frac{f(1)}{1} = 1$ (we must have $f(0) = 0$ and $f(1) = 1$ because and $f$ is a bijection and $f'(x) \geq 0$ so $f$ is also monotone increasing), i.e. $f(x) \geq x$. Therefore $f^{-1}(x) \leq x$ so that (using our subclaim again) $$ \frac{x}{f^{-1}(x)} =\frac{f(f^{-1}(x))}{f^{-1}(x)} \leq \frac{f(x)}{x}, $$ as desired.
Now to show that $\frac{f(x_1)}{x_1} \geq \frac{f(x_2)}{x_2}$ whenever $x_1 \leq x_2$, observe that it is equivalent to show that the derivative of the function $\frac{f(x)}{x}$ is nonpositive on $(0, 1)$, i.e. $\frac{f(x)}{x}$ is monotone decreasing. We can calculate directly that $$ \frac{d}{d x} \left(\frac{f(x)}{x}\right) = \frac{f'(x)}{x} - \frac{f(x)}{x^2}. $$
Finally we claim that $f'(x) \leq \frac{f(x)}{x}$ for all $x \in (0, 1)$. Assuming this for the moment the last equation becomes the inequality $$ \frac{d}{d x} \left(\frac{f(x)}{x}\right) = \frac{f'(x)}{x} - \frac{f(x)}{x^2} \leq \frac{f(x)}{x^2} - \frac{f(x)}{x^2} = 0, $$ as desired.
So we must finally show that $f'(x) \leq \frac{f(x)}{x}$ or equivalently $f(x) - x f'(x) \geq 0$ for all $x \in (0, 1)$. Because the LHS $f(x) - x f'(x)$ is $0$ when $x = 0$, it is in turn enough to show that the derivative of $f(x) - x f'(x)$ is nonnegative on $(0, 1)$. We can just compute that $$ \frac{d}{dx} (f(x) - x f'(x)) = f'(x) - f'(x) - x f''(x) = - x f''(x). $$ By hypothesis $f''(x) \leq 0$, thus $-x f''(x) \geq 0$ on $(0, 1)$, and so this completes the proof.