74th Romanian Mathematical Olympiad

Let $g:\mathbb{R}\to \mathbb{R}$ defined by $g(x)=x+\sin (x)$, for any $x\in \mathbb{R}$. Obviously, $g$ is differentiable on $\mathbb{R}$, with $g^{\prime }(x)>0$, for all $x\in ((2k-1)\pi ,(2k+1)\pi )$, for $k\in \mathbb{Z}$. Thus $g$ is increasing on any interval $[(2k-1)\pi ,(2k+1)\pi ]$, with $k\in \mathbb{Z}$, so $g$ is increasing on $\mathbb{R}$. As ${\lim }_{x\to -\infty }g(x)=-\infty$ and ${\lim }_{x\to \infty }g(x)=\infty$, $g$ is one-to-one, with $g^{-1}:\mathbb{R}\to \mathbb{R}$ continuous and increasing on $\mathbb{R}$. The given inequality can thus be written $$g(f(x))\ge x,\text{ for any }x\in \mathbb{R}⟺f(x)\ge g^{-1}(x),\text{ for any }x\in \mathbb{R}.$$ Then $${\int }_0^{\pi }f(x)dx\ge {\int }_0^{\pi }{g}^{-1}(x)dx.$$ As $g(0)=0$ and $g(\pi )=\pi$, by Young inequality, we have $$\begin{array}{rl}{\int }_0^{\pi }{g}^{-1}(x)dx& =\pi \cdot g(\pi )-0\cdot g(0)-{\int }_0^{\pi }g(x)dx\\ & ={\pi }^2-\left(\frac{{\pi }^2}{2}-\cos (\pi )\right)+(0-\cos (0))=\frac{{\pi }^2}{2}-2\end{array}$$ which gives the result.