67th Romanian Mathematical Olympiad

(a) To establish the required inequality write: $$\begin{array}{rl}{\int }_0^2\frac{f(x+1)}{f(x)}dx& ={\int }_0^1\frac{f(x+1)}{f(x)}dx+{\int }_1^2\frac{f(x+1)}{f(x)}dx\\ & ={\int }_0^1\frac{f(x+1)}{f(x)}dx+{\int }_1^2\frac{f(x-1+2)}{f(x-1+1)}dx\\ & ={\int }_0^1\frac{f(x+1)}{f(x)}dx+{\int }_0^1\frac{f(x+2)}{f(x+1)}dx\\ & ={\int }_0^1\left(\frac{f(x+1)}{f(x)}+\frac{f(x)}{f(x+1)}\right)dx\\ & \ge {\int }_0^{1}2dx=2.\end{array}$$

(b) If $1$ is a period of $f$, then ${\int }_0^2\frac{f(x+1)}{f(x)}dx={\int }_0^{2}dx=2$. Conversely, if ${\int }_0^2\frac{f(x+1)}{f(x)}dx=2$, then $${\int }_0^1{\left(\sqrt{\frac{f(x+1)}{f(x)}}-\sqrt{\frac{f(x)}{f(x+1)}}\right)}^{2}dx=0,$$ so $\sqrt{\frac{f(x+1)}{f(x)}}=\sqrt{\frac{f(x)}{f(x+1)}}$, $0\le x\le 1$, by continuity, and $f(x+1)=f(x)$, for all $x\in [0,1]$. If $1\le x\le 2$, then $f(x)=f((x-1)+1)=f(x-1)=f((x-1)+2)=f(x+1)$, so $f(x+1)=f(x)$ for all $x\in [0,2]$. Finally, if $x$ is any real number, then $2n\le x<2n+2$ for some integer $n$, and $f(x)=f(x-2n)=f((x-2n)+1)=f((x+1)-2n)=f(x+1)$. Consequently, $1$ is a period of $f$.