Upper bound on an integral

Question

Let $ f : [0,1] \to \mathbb{R} $ be a function satisfying: 1.) $ |f(x)| \leqslant a $ for some $ a < 1 $, and 2.) $ \int_0^1 f(x) dx = 0 $. I would like to know whether the following inequality holds: $$ \int_0^1 \frac{1}{1+f(x)} dx \stackrel{?}{\leqslant} \frac{1}{1-a^2} . $$ I've confirmed it numerically in some special cases, e.g., $ f(x) = a \cos(2\pi n x) $ for various $ a, n $. It is also easy to see that it holds (with equality) when $ f(x) = \pm a $, each on one half of the domain.

Answer 1

Since $|f(x)| \leq a$, then one has $$f(x)^2 \leq a^2, \quad \quad \text{so }\quad\quad 1-f(x)^2 \geq 1-a^2$$

Dividing by $(1+f(x))(1+a)$, one gets $$\dfrac{1-f(x)}{1+a} \geq \dfrac{1-a}{1+f(x)}$$

Now integrate between $0$ and $1$ : using the assumption $\displaystyle \int_0^1 f(x) dx = 0$, you get $$(1-a)\int_0^1 \dfrac{dx}{1+f(x)} \leq \dfrac{1}{1+a}\int_0^1 (1-f(x)) dx = \dfrac{1}{1+a}$$

so finally $$\boxed{\int_0^1 \dfrac{dx}{1+f(x)} \leq \dfrac{1}{1-a^2}}$$