TITLE: How to solve the following integral inequality $(\alpha+2)\cdot \int_0^1 x^\alpha(f(x)+f^{-1}(x)) \, dx \le 2$
QUESTION [6 upvotes]: Let $f:[0, 1]\to[0, 1]$ be continuous and bijective with $f(0) = 0$. Prove that for every $\alpha \ge 0$: $$(\alpha+2)\cdot \int_0^1 x^\alpha (f(x)+f^{-1}(x)) \, dx \le 2$$
To be fair I don't even know how to begin solving it. I'm pretty sure $f(0)=0$ should help me but I'm not sure how
Solution:
$f$ continuous and bijective with $f(0) = 0$ and $f(1)=1$$\implies$ $f$ is increasing
Using Young: $\int_0^x f(t) \, dt+\int_0^x f^{-1}(t) \, dt \ge x^2, x\in[0, 1]$ if and only if $f(x)=x$; so we can rewrite it as $\int_0^x (f(t)+f^{-1}(t)-2t) \, dt\ge 0, x\in[0, 1]$
Let $G:[0, 1]\to \mathbb{R}$ with $G(x)=\int_0^x (f(t)+f^{-1}(t)-2t) \, dt$. Then $G(1)=G(0)=0, G(x)\ge 0, x\in[0, 1]$
$G$ differentiable and for every $\alpha\ge 0$:
$$\int_0^1 x^{\alpha}(f(x)+f^{-1}(x)) \, dx = 2\int_0^1 x^{\alpha+1} \, dx + \int_0^1 x^\alpha(f(x)+f^{-1}(x)-2x) \, dx = \frac{2}{\alpha+2} + \int_0^1 x^\alpha G'(x)\,dx = \frac{2}{\alpha+2}+\lim_{a \to0} (G(1)-G(a)\cdot a^\alpha -\int_a^1 \alpha\cdot x^{\alpha-1} G(x) \, dx) \le\frac{2}{\alpha+2}$$ therefore $$(\alpha+2)\cdot \int_0^1 x^\alpha (f(x)+f^{-1}(x)) \, dx \le 2$$

REPLY [4 votes]: Let $\Phi(x)=\int^x_0(f+f^{-1})$. We have that $\Phi(x)\geq x^2$ and $\Phi(1)=1$. Integration by parts gives
$$\begin{align}
\int^1_0x^\alpha (f^(x)+f^{-1}(x))\,dx&= \Phi(x)x^\alpha\Big|^1_0-\alpha\int^1_0 x^{\alpha-1}\Phi(x)\,dx\\
&\leq 1-\alpha\int^1_0x^{\alpha+1}\,dx=1-\frac{\alpha}{\alpha+2}=\frac{2}{\alpha+2}
\end{align}
$$
The conclusion follows.