1. Given Information and Setup:
- Let $f: \mathbb{R} \to \mathbb{R}^+$ be a continuous and periodic function with period $T$.
- We need to prove that for all $\alpha \in \mathbb{R}$, the inequality $\int_0^T \frac{f(x)}{f(x+\alpha)} \, dx \ge T$ holds.

2. Using the Arithmetic Mean-Geometric Mean (AM-GM) Inequality:
- The AM-GM inequality states that for any non-negative functions $g(x)$ over an interval $[a, b]$,
$$\frac{1}{b-a} \int_a^b g(x) \, dx \ge \exp\left( \frac{1}{b-a} \int_a^b \ln g(x) \, dx \right).$$
- In our case, let $g(x) = \frac{f(x)}{f(x+\alpha)}$. Then,
$$\frac{1}{T} \int_0^T \frac{f(x)}{f(x+\alpha)} \, dx \ge \exp\left( \frac{1}{T} \int_0^T \ln \left( \frac{f(x)}{f(x+\alpha)} \right) \, dx \right).$$

3. Simplifying the Exponential Term:
- Notice that
$$\ln \left( \frac{f(x)}{f(x+\alpha)} \right) = \ln f(x) - \ln f(x+\alpha).$$
- Therefore,
$$\frac{1}{T} \int_0^T \ln \left( \frac{f(x)}{f(x+\alpha)} \right) \, dx = \frac{1}{T} \int_0^T \ln f(x) \, dx - \frac{1}{T} \int_0^T \ln f(x+\alpha) \, dx.$$

4. **Using the Periodicity of $f$:**
- Since $f$ is periodic with period $T$, the integral of $\ln f(x)$ over one period is the same as the integral of $\ln f(x+\alpha)$ over the same period:
$$\int_0^T \ln f(x) \, dx = \int_0^T \ln f(x+\alpha) \, dx.$$
- Thus,
$$\frac{1}{T} \int_0^T \ln f(x) \, dx - \frac{1}{T} \int_0^T \ln f(x+\alpha) \, dx = 0.$$
- This implies that
$$\exp\left( \frac{1}{T} \int_0^T \ln \left( \frac{f(x)}{f(x+\alpha)} \right) \, dx \right) = \exp(0) = 1.$$

5. Applying the AM-GM Inequality:
- From the AM-GM inequality, we have
$$\frac{1}{T} \int_0^T \frac{f(x)}{f(x+\alpha)} \, dx \ge 1.$$
- Multiplying both sides by $T$, we get
$$\int_0^T \frac{f(x)}{f(x+\alpha)} \, dx \ge T.$$

6. Equality Condition:
- Equality holds if and only if $\frac{f(x)}{f(x+\alpha)}$ is constant for almost every $x$. Given the continuity of $f$, this implies $f(x) = f(x+\alpha)$ for all $x$.

$\blacksquare$