Integral inequality on a periodic function Given $f:\mathbb R\to \mathbb R^+$ continuous and periodic of period $T\geq 0$, I am asked to prove that $$\int_0^T\frac{f(x)}{f(x+\alpha)}\mathrm dx\geq T,\; \forall \alpha\in\mathbb R.$$ How does one tackle this problem? It seems absolutely non trivial to me. 
 A: Write $f(x):=e^{g(x)}$. Then
$$\int_0^T{f(x)\over f(x+\alpha)}\ dx=T \int_0^T e^{g(x)-g(x+\alpha)}\ {dx\over T}\ ,$$
and the convexity of the exponential function implies that this is
$$\geq T \exp\left(\int_0^T \bigl(g(x)-g(x+\alpha)\bigr)\ {dx\over T}\right) =T e^0=T\ .$$
A: Here's a solution that came to my mind.
Firstly, we prove the following (easy) consequence of $g(x) = \frac{f(x)}{f(x+\alpha)}$ having period $T$: $$\int_{0}^{nT} g(x) \ dx = n \int_{0}^{T} g(x) \ dx$$ for any natural number $n$.
It is easy enough to see that we may take $\alpha > 0$ (Replace $\alpha$ by $\alpha +kT$ for some $k \in \mathbb N$). We first handle the case where $\alpha = \frac nm T$ for some positive integers $n,m$. Now, $$\int_{0}^{nT} g(x) \ dx = \sum_{j=0}^{m-1} \int_{\frac jmnT}^{\frac{(j+1)}mnT} g(x) \ dx = \sum_{j=0}^{m-1} \int_{j\alpha}^{(j+1)\alpha} \frac{f(x)}{f(x+\alpha)} \ dx = \sum_{j=0}^{m-1} \int_{0}^{\alpha} \frac{f(x+j\alpha)}{f(x+(j+1)\alpha)} \ dx$$
The left hand side is $\displaystyle  n \int_{0}^{T} g(x) \ dx$ by our initial observation, also, the summation can be brought inside the integral on the right hand side, to get
$$ n \int_{0}^{T} g(x) \ dx = \int_{0}^{\alpha} \sum_{j=0}^{m-1} \frac{f(x+j\alpha)}{f(x+(j+1)\alpha)} \ dx$$
Now, we have the following due to the AM-GM inequality: $$\frac{a_0}{a_1} + \frac{a_1}{a_2} + \cdots + \frac{a_{m-1}}{a_0} \ge m$$ therefore, the term inside the integral is $\ge m$, so the integral itself is $\ge m \alpha = nT$, and so we're done.
Moving on to irrational $T/\alpha$. Then we have the sequence of fractional parts $\{n T/\alpha \}$ is dense in the unit interval $(0,1)$, therefore, we choose arbitrary large $n$ such that $\{ nT/\alpha \} < \epsilon$, for some $\epsilon >0$. So, we have that if $\lfloor n \alpha/T\rfloor = m$, then $m \alpha < nT < m\alpha + \epsilon \alpha$. Thus, $$n \int_{0}^{T} g(x) \ dx \ge m \alpha \ge nT - \epsilon \alpha \implies \int_{0}^{T} g(x) \ dx \ge T - \epsilon \alpha /n$$ by the same process as before (we just need to add that $f(m \alpha) < f(nT) + \delta $ where $\delta \rightarrow 0$ as $\epsilon \rightarrow 0$) and since $\epsilon$ can be made arbitrarily small, therefore, we're done.
A: Well, one should organize his mind I think and proceed by steps. When you deal with a periodic function, the first thing you should keep in mind is that, for any real number $p$, $$\int_0^Tf(x)\mathrm d x=\int_0^Tf(x+p)\mathrm dx;$$ it is not difficult to prove so try do it yourself.
The next step is quite tricky (I admit I've already seen something similar some time ago) and is based on the following observation. Suppose the thesis to be true for $2\alpha\in\mathbb R$ and for any $f$ positive, continuous and periodic of period $T$; then it holds also for $\alpha$ and for any $f$ positive, continuous and periodic of period $T$. The proof of this fact goes as follows: $$T\leq\int_0^T \sqrt{\frac{f(x)}{f(x+2\alpha)}}\mathrm d x=\int_0^T\sqrt{\frac{f(x)}{f(x+\alpha)}\frac{f(x+\alpha)}{f(x+2\alpha)}}\mathrm dx;$$ by Cuachy Schwarz the latter term is lesser than or equal to $$\sqrt{\int_0^T\frac{f(x)}{f(x+\alpha)}\mathrm dx\int_0^T\frac{f(x+\alpha)}{f(x+2\alpha)}\mathrm dx}=\int_0^T \frac{f(x)}{f(x+\alpha)} \mathrm dx.$$ Remark We have use essentially that the functions considered are positive, for otherwise there could be problems with the square roots and with Cauchy Schwarz.
Now we observe that, for any positive, periodic of period $T$ and continuous function the thesis is true for $\alpha=nT$, with $n\in\mathbb Z$. Then an easy induction proves that the result is true for all the reals of the form $$\alpha=T\frac{n}{2^k}, \; n\in\mathbb Z, k\in\mathbb N.$$ Then we are done, since these numbers are dense in $\mathbb R$.
