TITLE: Integral inequality on a periodic function
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: 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.