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Prove the inequality

lfdahl

Well-known member
Nov 26, 2013
719
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:

$$\int_{0}^{1}\frac{f(x)}{f(x+\frac{1}{2})} \,dx \geq 1.$$
 

Olinguito

Well-known member
Apr 22, 2018
251
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.
 

lfdahl

Well-known member
Nov 26, 2013
719
Thankyou, Olinguito , for your participation and correct answer. Clever done!

An alternative solution:

\[\int_{0}^{1}\frac{f(x)}{f(x+\frac{1}{2})}dx = \int_{0}^{\frac{1}{2}}\frac{f(x)}{f(x+\frac{1}{2})}dx + \int_{\frac{1}{2}}^{1}\frac{f(x)}{f(x+\frac{1}{2})}dx \\ = \int_{0}^{\frac{1}{2}}\frac{f(x)}{f(x+\frac{1}{2})}dx+\int_{0}^{\frac{1}{2}}\frac{f(x+\frac{1}{2})}{f(x+1)}dx \\= \int_{0}^{\frac{1}{2}}\frac{f(x)}{f(x+\frac{1}{2})}dx+\int_{0}^{\frac{1}{2}}\frac{f(x+\frac{1}{2})}{f(x)}dx \\= \int_{0}^{\frac{1}{2}}\left ( \frac{f(x)}{f(x+\frac{1}{2})}+\frac{f(x+\frac{1}{2})}{f(x)} \right )dx \\
\geq \int_{0}^{\frac{1}{2}} 2\sqrt{\frac{f(x)}{f(x+\frac{1}{2})}\cdot \frac{f(x+\frac{1}{2})}{f(x)} }dx \\=2\int_{0}^{\frac{1}{2}}dx = 1.\]
 

MountEvariste

Well-known member
Jun 29, 2017
79
Beautiful solution! I observed this symmetry but didn't reach the step where you achieve the inequality.
 

Olinguito

Well-known member
Apr 22, 2018
251
Beautiful solution! I observed this symmetry but didn't reach the step where you achieve the inequality.
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 $\left(\sqrt X-\sqrt Y\right)^2\ge0$.
 

MountEvariste

Well-known member
Jun 29, 2017
79
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 $\left(\sqrt X-\sqrt Y\right)^2\ge0$.
I know. I was talking about my own attempt to solve the problem prior to seeing lfdahl 's solution.
 

lfdahl

Well-known member
Nov 26, 2013
719
I know. I was talking about my own attempt to solve the problem prior to seeing lfdahl 's solution.
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.