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:

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

Solution

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

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

Thus:

\begin{align*}
\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}u \\
 &=\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{align*}

since it holds that $x + \frac{1}{x} \geq 2$ forall $x>0$.