Show that $\int_0^1[1+f(x)]dx\int_0^1\frac{1}{1+f(x)}dx\le1.125$

Question

Show that $$\int_0^1[1+f(x)]dx\int_0^1\frac{1}{1+f(x)}dx\le1.125$$ where $f:[0,1]\rightarrow [0,1]$.

In the beginning, I was going to use Chebyshev's inequality but then noticed that we don't have any info about monotonicity :) Now I don't know how to do it. Any hint is appreciated.

Answer 1

Note that since $0 \le f(x) \le 1$ in $[0,1]$, denoting $1 \le g(x) := 1 + f(x) \le 2$ $$\left(2 - g(x)\right)\left(\frac{1}{g(x)} - 1\right) \le 0 \implies \frac{2}{g(x)} + g(x) \le 3$$

Integrating both sides \begin{align*} 2\int_0^1 \frac{1}{g(x)}\,dx + \int_0^1 g(x)\,dx \le 3 \end{align*}

By AM-GM inequality $$2\left(\int_0^1 \frac{1}{g(x)}\,dx\right) \left( \int_0^1 g(x)\,dx\right) \le \frac{1}{4}\left(2\int_0^1 \frac{1}{g(x)}\,dx + \int_0^1 g(x)\,dx\right)^2 \le \frac{9}{4}$$ proving the claim.

Answer 2

Hint. Show that $f([0,1])\subseteq [0,1]$ implies $$\frac{1}{2}\left((1+f(x))+\frac{2}{1+f(x)}\right)\leq \frac{3}{2}.$$