Integral inequality with differentiable function

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Integral inequality with differentiable function

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Source: Romania NMO 2025 12.2

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#1 h Apr 2, 2025, 2:29 PM

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Let $f \colon [0,1] \to \mathbb{R}$ be a differentiable function such that its derivative is an integrable function on $[0,1]$ , and $f(1)=0$ . Prove that $\int_0^1 (xf'(x))^2 dx \geq 12 \cdot \left( \int_0^1 xf(x) dx\right)^2$

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#2 h Apr 2, 2025, 3:35 PM • 1 Y

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Using IBP, $\displaystyle \int\textstyle xf=\displaystyle \int\textstyle \frac{-x^2f'}2$ ,
$[asy]usepackage("amsmath, amssymb, tikz, tikz-cd"); label("\begin{tikzcd}[ampersand replacement=\&] \displaystyle\int\scriptstyle x^2\displaystyle\int \scriptstyle(xf')^2\ \ge\ \left(\displaystyle\int\scriptstyle x^2f'\right)^2\&\&\displaystyle\int \scriptstyle(xf')^2\ \ge\ 3\left(\displaystyle\int\scriptstyle x^2f'\right)^2 \arrow[Rightarrow, from=1-1, to=1-3] \end{tikzcd}");[/asy]$
so
$[asy]usepackage("amsmath, amssymb, tikz, tikz-cd"); label("\begin{tikzcd}[ampersand replacement=\&] \hspace{1pt}\&\&\color{blue}\displaystyle\int \scriptstyle(xf')^2\ \ge\ 12\left(\displaystyle\int\scriptstyle xf\right)^2. \arrow[Rightarrow, blue, from=1-1, to=1-3] \end{tikzcd}");[/asy]$

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#3 h Apr 4, 2025, 9:33 PM

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Easy one...

Note that using integration by parts, we get $\int_0^1 xf(x) \ dx = \left[f(x)\frac{x^2}{2}\right]_0^1 - \int_0^1 f'(x)\frac{x^2}{2} \ dx=-\int_0^1 f'(x)\frac{x^2}{2} \ dx$ So, we need to show
$\int_0^1 (xf'(x))^2 \ dx\ge 12 \left(\int_0^1 xf(x) \ dx\right)^2=12\left(\int_0^1 f'(x)\frac{x^2}{2} \ dx\right)^2=3\left(\int_0^1 x^2f'(x) \ dx\right)^2$ $\iff \frac{1}{3}\left(\int_0^1 (xf'(x))^2 \ dx\right)\ge \left(\int_0^1 x^2f'(x) \ dx\right)^2$ $\iff \left(\int_0^1 x^2 \ dx\right)\left(\int_0^1 (xf'(x))^2 \ dx\right)\ge \left(\int_0^1 x^2f'(x) \ dx\right)^2$ which is just Cauchy Schwarz inequality.

https://external-preview.redd.it/Vt8un6DvXelUjrQPqHsJXaIijhQIMDRU50RjKVXe2JM.jpg?auto=webp&s=fbd4da7e4d2893906b86cebbff64976f0e1c03e2

@below, thanks for pointing out the typo.

This post has been edited 1 time. Last edited by Fibonacci_math, Apr 9, 2025, 10:08 PM

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#4 h Apr 9, 2025, 9:37 PM

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same solution as the previous two
applying IBP on $\int_{0}^{1}xf$
we get
$\int_{0}^{1} xf=\displaystyle \int_{0}^{1}\textstyle \frac{-x^2f'}2$

now it's equivalent to proof that
$\int_0^1 (xf'(x))^2 dx \geq 12 \cdot \left(\int_{0}^{1}\textstyle \frac{-x^2f'}2\right)^2$
$\iff \frac{1}{3} \int_0^1 (x f'(x))^2 \, dx \geq \left( \int_0^1 x^2 f'(x) \, dx \right)^2$
$\iff \int_{0}^{1}x^{2} \int_0^1 (x f'(x))^2 \, dx \geq \left( \int_0^1 x^2 f'(x) \, dx \right)^2$
which is CS

https://static.wikia.nocookie.net/minecraft_gamepedia/images/e/eb/Plains_Baby_Villager_Base_JE2.png/revision/latest/scale-to-width/360?cb=20220612221105

$\mathbb{QED}$ $\blacksquare$
@above i think u made a typo by not putting $^{2}$ on the $\text{LHS}$ its a typo yes but that messes up the CS inequality

This post has been edited 4 times. Last edited by Levieee, Apr 9, 2025, 9:44 PM

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