Dan Sitaru's Integral Inequality with Powers of a Function

Problem

Dan Sitaru's Integral Inequality with Powers of a Function

Solution 1

Using repeatedly the Cauchy-Schwarz inequality,

$\displaystyle \begin{align} &\left(\int_0^1f^5(x)dx\right)\left(\int_0^1f^7(x)dx\right)\left(\int_0^1f^9(x)dx\right)\left(\int_0^1f^3(x)dx\right)\\ &\qquad\qquad\qquad=\int_0^1\left(f^2\sqrt{f(x)}\right)^2dx\cdot\int_0^1\left(f^3\sqrt{f(x)}\right)^2dx\,\cdot\\ &\qquad\qquad\qquad\qquad\qquad\cdot\int_0^1\left(f^4\sqrt{f(x)}\right)^2dx\cdot\int_0^1\left(f\sqrt{f(x)}\right)^2dx\\ &\qquad\qquad\qquad\ge \left(\int_0^1f^6(x)dx\right)^2\cdot\left(\int_0^1f^6(x)dx\right)^2\\ &\qquad\qquad\qquad = \left[\left(\int_0^1f^6(x)dx\right)\cdot\left(\int_0^11^2dx\right)\right]^4\\ &\qquad\qquad\qquad\ge\left(\int_0^1f^3(x)dx\right)^8=\sqrt[7]{2^8}, \end{align}$

so that

$\displaystyle \sqrt[7]{2}\left(\int_0^1f^5(x)dx\right)\left(\int_0^1f^7(x)dx\right)\left(\int_0^1f^9(x)dx\right)\ge\sqrt[7]{2^8}.$

In other words,

$\displaystyle \left(\int_0^1f^5(x)dx\right)\left(\int_0^1f^7(x)dx\right)\left(\int_0^1f^9(x)dx\right)\ge 2.$

Equality is attained for $f(x)\equiv\sqrt[21]{2}.$

Solution 2

By Hölder's inequality,

$\displaystyle\left(\int_0^1f^n(x)dx)\right)^{\frac{3}{n}}\left(\int_0^1dx)\right)^{\frac{n-3}{2n}}\left(\int_0^1dx)\right)^{\frac{n-3}{2n}}\ge\int_0^1f^3(x)dx=\sqrt[7]{2},$

implying, in particular,

$\displaystyle \begin{align} \int_0^1f^5(x)dx)\ge \sqrt[21]{2^5},\\ \int_0^1f^7(x)dx)\ge \sqrt[21]{2^7},\\ \int_0^1f^9(x)dx)\ge \sqrt[21]{2^9}. \end{align}$

Multiplying the three we have

$\displaystyle\left(\int_0^1f^5(x)dx\right)\left(\int_0^1f^7(x)dx\right)\left(\int_0^1f^9(x)dx\right)\ge\sqrt[21]{2^{5+7+9}}=2.$

Solution 3

Consider the convex function $g(x)=x^k\,$ with $k>1.\,$ Applying Jensen's inequality (noting that powers of f are non-negative and Riemann-integrable),

$\displaystyle \begin{align} g\left(\int_0^1 f^3(x) dx\right) &\leq \int_0^1 g\left(f^3(x)\right)dx \\ \Longrightarrow \left(\int_0^1 f^3(x) dx\right)^k &\leq \int_0^1 f^{3k}(x) dx \end{align}$

Multiplying the inequalities for $k=5/3,~7/3,~9/3$ together,

$\displaystyle 2 = \left(\int_0^1 f^3(x) dx\right)^7 \leq\left(\int_0^1 f^5(x)dx\right) \left(\int_0^1 f^7(x)dx\right) \left(\int_0^1 f^9(x)dx\right).$

Solution 4

By Hölder's inequality,

$\displaystyle \left(\int_0^1f^{3p}(x)dx\right)^{\frac{1}{p}}\left(\int_0^11^qdx\right)^{\frac{1}{q}}\ge \int_o^1f^3(x)dx,$

Hence, $\displaystyle \int_0^1f^{3p}(x)dx\ge \left(\sqrt[7]{2}\right)^p.$

Allora, taking $\displaystyle p=\{\frac{5}{3},\frac{7}{3},\frac{9}{3}\}\,$ and merging, we get the result.

Acknowledgment

This problem from the Romanian Mathematical Magazine has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. He later mailed his solution (Solution 1) in a LaTeX file, along with a solution (Solution 2) by Chris Kyriazis (Greece). Solution 3 is by Amit Itagi; Solution 4 is by N. N. Taleb.

 

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