List questions from integral-inequality

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Prove an integration inequality $\int_{0}^{1}\,\left(\int_{0}^{x}\,f(y)\,{\rm d}y\right)^2{\rm d}x\leqslant -\frac{mM}{6(M-m)^2}(3m^2-8mM+3M^2)$

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Probabilistic Proof of Chebyshev Inequality - Step 3

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Integral inequality $\int_{0}^{1}f^2\left(x\right)dx\geq\frac{3}{13}$

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Integral inequality $e^{\int _0^1f\left(x\right)dx-1\:}\le \frac{1}{4}\int _0^1\left(x+1\right)e^{f\left(x\right)}dx$

Published on 18 Apr 2026 - 16:00

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An inequality with Sobolev norm.

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Divergence of a Sequence of Integrals - An Olympiad Problem

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274

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Integral inequality between function and derivative

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Average integration is increasing with measure

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Interpolation like inequality ,Question from Brezis' book exercise 8.15

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Geometric interpretation of Hölder's inequality

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Explanation for the proof of Minkowski's inequality, Theorem 3.5, Rudin's RCA

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Inequality and integrals

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103

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$L^2$ convergence and pointwise-norm

Published on 29 Apr 2026 - 12:15

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$\int_0^s \log f(t)dt \le \int_0^s \log g(t)dt$ for every $s\in (0,1)$ implies that $\int_0^1 \log (f(t)+1)dt \le \int_0^1 \log (g(t)+1)dt$?

Published on 29 Apr 2026 - 12:54

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converge in $\mu$ almost everywhere

Prove an integration inequality $\int_{0}^{1}\,\left(\int_{0}^{x}\,f(y)\,{\rm d}y\right)^2{\rm d}x\leqslant -\frac{mM}{6(M-m)^2}(3m^2-8mM+3M^2)$

Probabilistic Proof of Chebyshev Inequality - Step 3

Integral inequality $\int_{0}^{1}f^2\left(x\right)dx\geq\frac{3}{13}$

Integral inequality $e^{\int _0^1f\left(x\right)dx-1\:}\le \frac{1}{4}\int _0^1\left(x+1\right)e^{f\left(x\right)}dx$

An inequality with Sobolev norm.

Divergence of a Sequence of Integrals - An Olympiad Problem

Integral inequality between function and derivative

Average integration is increasing with measure

Interpolation like inequality ,Question from Brezis' book exercise 8.15

Geometric interpretation of Hölder's inequality

Explanation for the proof of Minkowski's inequality, Theorem 3.5, Rudin's RCA

Inequality and integrals

$L^2$ convergence and pointwise-norm

$\int_0^s \log f(t)dt \le \int_0^s \log g(t)dt$ for every $s\in (0,1)$ implies that $\int_0^1 \log (f(t)+1)dt \le \int_0^1 \log (g(t)+1)dt$?

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