Integral and Inequality

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a positive real valued and continuous function such that it is periodic of period $T=1$ . Prove that

$\displaystyle \int_0^1 \frac{f(x)}{f \left(x + \frac{1}{2} \right)}\, {\rm d}x \geq 1$

Solution

Since the function $f$ is $1$ periodic , then it holds that:

$f(x+1)=f(x)=f(x-1) \quad \text{forall} \; x \in \mathbb{R}$

Thus:

$\begin{aligned} \int_{0}^{1}\frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x &=\int_{0}^{1/2} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x + \int_{1/2}^{1} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x \\ &= \int_{0}^{1/2} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x + \int_{1/2}^{1} \frac{f(x)}{f\left ( x+1-1 + \frac{1}{2} \right )} \, {\rm d}x\\ &= \int_{0}^{1/2} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x + \int_{1/2}^{1} \frac{f(x)}{f\left ( x- \frac{1}{2} \right )} \, {\rm d}x \\ &\!\!\!\!\!\overset{u=x-1/2}{=\! =\! =\! =\! =\!} \int_{0}^{1/2} \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} \, {\rm d}x + \int_{0}^{1/2} \frac{f\left ( x + \frac{1}{2} \right )}{f(x)} \, {\rm d}x \\ &=\int_{0}^{1/2} \left [ \frac{f(x)}{f\left ( x+\frac{1}{2} \right )} + \frac{f\left ( x+\frac{1}{2} \right )}{f(x)} \right ] \, {\rm d}x \\ &\geq \int_{0}^{1/2} 2 \, {\rm d}x \\ &=1 \end{aligned}$