Show that $\int_0^T\frac{f(x)}{f(x+\alpha)}dx\ge T$ I'm struggling with the following problem, any hint will be appreciated.

The function $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ is continue and periodic with period $T$ show that: 
  $$\forall\alpha\in\mathbb{R}, \int_0^T\frac{f(x)}{f(x+\alpha)}dx\ge T$$

 A: Hint: rewrite the inequality as
$$ \frac{1}{T}\int_0^T \frac{f(x)}{f(x+\alpha)} \mathrm{d}x \geq 1$$
and apply Jensen's inequality with the convex function $y \mapsto - \log (y) $. 

Observe that $-\log$ is a convex function. So Jensen's inequality for the probability measure $\frac{1}{T} \mathrm{d}x$ on $[0,T]$ (we normalize by $1/T$ to make the measure having total mass 1) implies
$$ - \log \left(\frac{1}{T} \int_0^T \frac{f(x)}{f(x+a)} \mathrm{d}x\right) \leq -\frac{1}{T} \int_0^T \log \frac{f(x)}{f(x+a)} \mathrm{d}x $$
The right hand side we use the fact that 
$$ \log a / b = \log a - \log b $$
to get 
$$ = - \frac{1}{T} \left( \int_0^T \log f(x) \mathrm{d}x - \int_0^T \log f(x+a) \mathrm{d}x \right) $$
Since $f$ is $T$-periodic, the integral of $\log f(x)$ and $\log f(x+a)$ over $[0,T]$ is the same. So the right hand side vanishes. 
This means that
$$ - \log\left( \frac{1}{T}\int_0^T \frac{f(x)}{f(x+a)} \mathrm{d}x \right) \leq 0 $$
or that 
$$ \log \left( \frac{1}{T} \int_0^T \frac{f(x)}{f(x+a)} \mathrm{d}x \right) \geq 0 $$
Now, the exponential function $y \mapsto e^y$ is strictly increasing, and so order preserving. So raising both sides to a power you recover
$$ \frac{1}{T}\int_0^T \frac{f(x)}{f(x+a)} \mathrm{d}x \geq e^0 = 1 $$
as desired. 
A: After trying a little bit harder I've found a solution.
First, note that $\forall\alpha\in\mathbb{R}$, $$\int_0^Tg(x)dx=\int_0^Tg(x+\alpha)dx \tag1$$ for $g$ a continue and periodic function with period $T$. 
The demonstration of $(1)$ is straightforward and only uses basic properties of integrals and the fact that the period of $g$ is T. Also note that this result is very intuitive because it represents the same integral just displaced $\alpha$ units.
Let $I=\int_0^T\frac{f(x)}{f(x+\alpha)}$, as $f$ is periodic then $\frac{f(x)}{f(x+\alpha)}$ must be periodic with the same period $T$. Applying the property $(1)$ we have:
$$I=\int_0^T\frac{f(x+(i-1)\alpha)}{f(x+i\alpha)}$$
for all $i\in\mathbb{N}$. Now, for $n$ a positive integer we have
$$I=\frac{nI}{n}=\frac{\sum_{i=1}^{n}\int_0^T\frac{f(x+(i-1)\alpha)}{f(x+i\alpha)}}{n}=\int_0^T\frac{\sum_{i=1}^n\frac{f(x+(i-1)\alpha)}{f(x+i\alpha)}}{n}\ge\int_0^T\sqrt[n]{\prod_{i=1}^n\frac{f(x+(i-1)\alpha)}{f(x+i\alpha)}}=\int_0^T\sqrt[n]{\frac{f(x)}{f(x+n\alpha)}}\ge\int_0^T\sqrt[n]{\frac{\min f(x)}{\max f(x)}}=T\sqrt[n]{\frac{\min f(x)}{\max f(x)}}$$
And now taking limits as $n\rightarrow \infty$ we get $I\ge T.$
