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

[lfdahl]

Let $f$ be a positive and continuous function on the real line which satisfies $f(x + 1) = f(x)$ for all numbers $x$.
Prove \[\int_{0}^{1}\frac{f(x)}{f(x+\frac{1}{2})}dx \geq 1.\]

 

[Euge]

Here is my solution.

Spoiler

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

 

[lfdahl]

Here is my solution.

Spoiler

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!