# 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.