MHB Calculus inequality challenge prove ∫10f(x)/f(x+1/2)dx≥1

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The discussion revolves around proving the inequality involving the integral of a positive, continuous, and periodic function. The specific inequality to prove is that the integral from 0 to 1 of the ratio of the function to its shifted version is greater than or equal to 1. Participants engage in sharing their solutions and insights on the problem. The periodic nature of the function, given by f(x + 1) = f(x), plays a crucial role in the proof. The conversation highlights the collaborative effort in tackling this calculus challenge.
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Let $f$ be a positive and continuous function on the real line which satisfies $f(x + 1) = f(x)$ for all numbers $x$.
Prove \[\int_{0}^{1}\frac{f(x)}{f(x+\frac{1}{2})}dx \geq 1.\]
 
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Here is my solution.

Note $\int_0^{1/2} \frac{f(x)}{f(x + 1/2)}\, dx = \int_{1/2}^1 \frac{f(x + 1/2)}{f(x)}\, dx$ and $$\int_{1/2}^1 \frac{f(x)}{f(x + 1/2)}\, dx = \int_0^1 \frac{f(x + 1/2)}{f(x + 1)}\, dx = \int_0^{1/2} \frac{f(x + 1/2)}{f(x)}\, dx$$ Therefore
$$\int_0^1 \frac{f(x)}{f(x+1/2)}\, dx = \int_0^1 \frac{f(x+1/2)}{f(x)}\, dx$$ and consequently
$$\int_0^1 \frac{f(x)}{f(x+1/2)}\, dx = \frac{1}{2}\int_0^1 \left(\frac{f(x)}{f(x+1/2)} + \frac{f(x+1/2)}{f(x)}\right)\, dx \ge \frac{1}{2}\int_0^1 2\sqrt{\frac{f(x)}{f(x+1/2)}\cdot \frac{f(x+1/2)}{f(x)}}\, dx = 1$$
 
Euge said:
Here is my solution.

Note $\int_0^{1/2} \frac{f(x)}{f(x + 1/2)}\, dx = \int_{1/2}^1 \frac{f(x + 1/2)}{f(x)}\, dx$ and $$\int_{1/2}^1 \frac{f(x)}{f(x + 1/2)}\, dx = \int_0^1 \frac{f(x + 1/2)}{f(x + 1)}\, dx = \int_0^{1/2} \frac{f(x + 1/2)}{f(x)}\, dx$$ Therefore
$$\int_0^1 \frac{f(x)}{f(x+1/2)}\, dx = \int_0^1 \frac{f(x+1/2)}{f(x)}\, dx$$ and consequently
$$\int_0^1 \frac{f(x)}{f(x+1/2)}\, dx = \frac{1}{2}\int_0^1 \left(\frac{f(x)}{f(x+1/2)} + \frac{f(x+1/2)}{f(x)}\right)\, dx \ge \frac{1}{2}\int_0^1 2\sqrt{\frac{f(x)}{f(x+1/2)}\cdot \frac{f(x+1/2)}{f(x)}}\, dx = 1$$

Excellent, Euge! (Nod) Thankyou very much for your participation!