Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(x)+\sin(f(x)) \ge x,$ for all $x \in \mathbb{R}.$ Prove that $$\int\limits_0^{\pi} f(x) \mathrm{d}x \ge \frac{\pi^2}{2}-2.$$
Joy Ma Tara The heading gives away the solution but lemme write that here:
Note that the function $x \rightarrow x+\sin x$ is an increasing function. Thus in order to minimize the integral, we need to look at the specific case when $f(x)+\sin f(x) = x$. Clearly $f$ is one one and as it's continuous it is strictly monotone. Hence $f^{-1}=g$ exists and given by $g(y)=y+\sin y$. Note that $g(0)=0$ and $g(\pi)=\pi$ by bijectivity of $g$. Therefore $$\int_0^{\pi}g(y) dy = \int_0^{\pi}(y+\sin y)dy = \frac{\pi^2}{2}+2$$Hence $$\min \left(\int_0^{\pi} f(x)dx \right) = \pi^2 - \int_0^{\pi}g(y) dy = \frac{\pi^2}{2}-2$$as desired.