MHB Prove Inequality: $f(x)$ Continuous Positive $\int_{0}^{1}$

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The discussion revolves around proving the inequality $$\int_{0}^{1}\frac{f(x)}{f(x+\frac{1}{2})} \,dx \geq 1$$ for a positive, continuous function $f$ that is periodic with a period of 1. Participants express appreciation for a solution provided by lfdahl, noting its cleverness and the symmetry observed in the problem. There is also clarification regarding the authorship of an alternative solution, emphasizing the importance of crediting original contributors. Overall, the conversation highlights various approaches to the problem and the community's engagement in mathematical discourse. The focus remains on the proof of the inequality involving the function $f$.
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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 the inequality:

$$\int_{0}^{1}\frac{f(x)}{f(x+\frac{1}{2})} \,dx \geq 1.$$
 
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By the Cauchy–Schwartz inequality for integrals,
$$\left(\int_0^1f(x)dx\right)^2\ \le\ \left(\int_0^1\frac{f(x)}{f\left(x+\frac12\right)}dx\right)^2\left(\int_0^1f\left(x+\frac12\right)dx\right)^2.$$
But $f(x)$ has period $1$ and so $\displaystyle\int_0^1f\left(x+\frac12\right)dx=\int_0^1f(x)dx$. The result follows.
 
Thankyou, Olinguito, for your participation and correct answer. Clever done!

An alternative solution:

\[\int_{0}^{1}\frac{f(x)}{f(x+\frac{1}{2})}dx = \int_{0}^{\frac{1}{2}}\frac{f(x)}{f(x+\frac{1}{2})}dx + \int_{\frac{1}{2}}^{1}\frac{f(x)}{f(x+\frac{1}{2})}dx \\ = \int_{0}^{\frac{1}{2}}\frac{f(x)}{f(x+\frac{1}{2})}dx+\int_{0}^{\frac{1}{2}}\frac{f(x+\frac{1}{2})}{f(x+1)}dx \\= \int_{0}^{\frac{1}{2}}\frac{f(x)}{f(x+\frac{1}{2})}dx+\int_{0}^{\frac{1}{2}}\frac{f(x+\frac{1}{2})}{f(x)}dx \\= \int_{0}^{\frac{1}{2}}\left ( \frac{f(x)}{f(x+\frac{1}{2})}+\frac{f(x+\frac{1}{2})}{f(x)} \right )dx \\
\geq \int_{0}^{\frac{1}{2}} 2\sqrt{\frac{f(x)}{f(x+\frac{1}{2})}\cdot \frac{f(x+\frac{1}{2})}{f(x)} }dx \\=2\int_{0}^{\frac{1}{2}}dx = 1.\]
 
lfdahl said:
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[sp]Beautiful solution! I observed this symmetry but didn't reach the step where you achieve the inequality.[/sp]
 
June29 said:
[sp]Beautiful solution! I observed this symmetry but didn't reach the step where you achieve the inequality.[/sp]
I believe lfdahl is using the fact that for any real, non-negative $X,Y$,
$$X+Y\ \ge\ 2\sqrt{XY}.$$
This comes from the AM–GM inequality, or alternatively from $\left(\sqrt X-\sqrt Y\right)^2\ge0$.
 
Olinguito said:
I believe lfdahl is using the fact that for any real, non-negative $X,Y$,
$$X+Y\ \ge\ 2\sqrt{XY}.$$
This comes from the AM–GM inequality, or alternatively from $\left(\sqrt X-\sqrt Y\right)^2\ge0$.
[sp]I know. I was talking about my own attempt to solve the problem prior to seeing lfdahl's solution.[/sp]
 
June29 said:
[sp]I know. I was talking about my own attempt to solve the problem prior to seeing lfdahl's solution.[/sp]

Please note: The alternative solution is from a 3rd party and not mine. Maybe, I should have pointed this out from the start. I´m sorry for the confusion.
 
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