An Inequality with Integrals and Radicals

Problem

An Inequality with Integrals and Radicals

Solution 1

By the AM-GM inequality,

$\displaystyle \sqrt[3]{f(x)}dx=\sqrt[3]{f(x)\cdot 1\cdot 1}\leq \frac{f(x)+1+1}{3}=\frac{f(x)+2}{3},$

so that

(1)

$\displaystyle \int_0^1 \sqrt[3]{f(x)}dx\leq \frac{1}{3}\int_0^1 f(x)dx+\frac{2}{3}\int_0^1 dx=1.$

Also

$\displaystyle \sqrt[5]{f(x)}dx=\sqrt[5]{f(x)\cdot 1\cdot 1}\leq \frac{f(x)+1+1+1+1}{5}=\frac{f(x)+4}{5},$

so that

(2)

$\displaystyle \int_0^1\sqrt[5]{f(x)}dx\leq \frac{1}{5}\int_0^1 f(x)dx+\frac{4}{5}\int_0^1 dx=1.$

Similarly,

(3)

$\displaystyle \int_0^1\sqrt[7]{f(x)}dx\leq \frac{1}{7}\int_0^1 f(x)dx+\frac{6}{7}\int_0^1 dx=1.$

By multiplying (1),(2),(3):

$\displaystyle \Bigr(\int_0^1 \sqrt[3]{f(x)}dx\Bigr)\Bigr(\int_0^1 \sqrt[5]{f(x)}dx\Bigr)\Bigr(\int_0^1 \sqrt[7]{f(x)}\Bigr)dx\leq 1.$

Solution 2

Function $x^n\,$ is convex for $x\ge 0,\,$ $n\,$ an integer so that, by the integral form of Jensen's inequality,

$\displaystyle\begin{align}\left(\int_0^1\sqrt[n]{f(x)}dx\right)^n &\le \int_0^1\left[\sqrt[n]{f(x)}\right]^ndx\\ &=\int_0^1f(x)dx=1. \end{align}$

It follows that each of the three integrals on the left is not greater than $1,\,$ and so is there product.

Solution 3

By the $L^p\text{-norm}\,$ inequality, in its genral form, for any $L^1\,$ function $f\,$ and for $0\lt p\le q,$

$\displaystyle \left(\int_a^bf(x)|^pdx\right)^{\frac{1}{p}}\ge\left(\int_a^b|f(x)|^qdx\right)^{\frac{1}{q}}.$

Here applay $\displaystyle p=\frac{1}{n_j},\,$ $q=1,\,$ $f(x)\ge 0,\,$ for all $n_j\gt 1:$

$\displaystyle \left(\int_a^bf(x)^{\frac{1}{n_j}}dx\right)^{n_j}\le 1.$

Thus,

$\displaystyle \prod_{j}\left(\int_a^bf(x)^{\frac{1}{n_j}}dx\right)^{n_j}\le 1.$

Solution 4

Using Hölder's inequality, with $\displaystyle p=n,\,$ $\displaystyle q=\frac{n}{n-1},\,$ $\displaystyle \frac{1}{p}+\frac{1}{q}=1,\,$ $\displaystyle \int_0^1ghdx\le\left(\int_0^1h^pdx\right)^{\frac{1}{p}}\left(\int_0^1g^qdx\right)^{\frac{1}{q}},\,$ $h=\sqrt[n]{f},\,$ $g\equiv 1:$

$\displaystyle\begin{align} \int_0^1\sqrt[n]{f(x)}dx&=\int_0^1\sqrt[n]{f(x)}\cdot 1dx\\ &\le\left(\int_0^1\left(\sqrt[n]{f(x)}\right)^ndx\right)^{\frac{1}{n}}\left(\int_0^11^{\frac{n}{n-1}}dx\right)^{\frac{n-1}{n}}\\ &=\left(\int_0^1f(x)dx)\right)^{\frac{1}{n}}=1. \end{align}$

Acknowledgment

The problem (from the Romanian Mathematical Magazine) has been posted by Dan Sitaru at the CutTheKnotMath facebook page, Dan later communicated by email his solution on a LaTeX file. The solution has been obtained independently by Chris Kyriazis. Solution 2 is by Amit Itagi; Solution 3 is by N. N. Taleb; Solution 4 is by Andrea Acquaviva.

 

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