Inegalitate integrala tot in gen Muirhead

[Cezar Lupu]

Fie \( f:[0,1]\to [0,\infty) \) o functie integrabila. Sa se arate ca

\( \int_0^1f(x)\cdot\int_0^1x^{3}f(x)dx\geq\int_0^1xf(x)dx\cdot\int_0^1x^{2}f(x)dx \).

Cezar Lupu & Mihai Piticari, Mathematical Reflections 2007


Observatie.
Desi aparent infonesiva, inegalitatea de mai sus chiar spune ceva:
In limbaj de statistica, daca \( X \) reprezinta o variabila aleatoare a repartitiei Pearson, atunci avem urmatoarea formula pentru media repartitiei mentionate mai sus:

\( M(X^{k})=\int_0^1x^{k}f(x)dx\forall k\geq 1 \),
iar daca \( k=0 \), atunci \( f \) este densitate de repartitie si in consecinta \( \int_0^1f(x)dx=1 \). Astfel, inegalitatea de mai sus (care are o rezolvare cu totul elementara) s-ar traduce in limbaj de repartitie Pearson astfel:

\( M(X)\cdot M(X^{2})\leq M(X^{3}) \).

[o.m.]

For any \( x,y\in [0;1] \) define

\( g(x,y)=f(x)f(y)(x^3+y^3-xy^2-x^2y)\geq 0 \).

Integrate on \( [0;1]^2 \) according to the variables \( x,y \) with Fubini and you will get this beautiful inequality.

[o.m.]

\( g(x,y)=f(x)f(y)(x+y)(x-y)^2\geq 0 \) on \( K=[0;1]^2 \)

\( \int\int_{K}(x^3f(x)f(y)+y^3f(x)f(y)-xy^2f(x)f(y)-x^2yf(x)f(y))dxdx\geq 0 \)

with Fubini

\( (\int_{0}^{1}x^3f(x)dx)(\int_{0}^{1}f(y)dy)+(\int_{0}^{1}f(x)dx)(\int_{0}^{1}y^3f(y)dy)\geq (\int_{0}^{1}xf(x)dx)(\int_{0}^{1}y^2f(y)dy)+(\int_{0}^{1}x^2f(x)dx)(\int_{0}^{1}yf(y)dy) \)

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In the title " Muirhead "
How use Muirhead inequality to write another proof ?

[Beniamin Bogosel]

Fubini asta rezolva orice problema???
Nu stiu teorema inca, dar de-abia astept sa o invat...

[Dragos Fratila]

Unde intervine repartitia aia Pearson? ca eu nu vad...

[aleph]

Fubini asta rezolva orice problema???
Nu stiu teorema inca, dar de-abia astept sa o invat...

Aici nu prea e nevoie de Fubini. Se integrează g în raport cu x pe [0,1] (privind y constant) şi apoi se integrează în raport cu y. Totul la nivel de clasa 12.

[o.m.]

Here another way with Riemann sums: expand g(x,y), choose for \( x=k/n \), \( y=q/m \) with k,q integers from 1 to n and 1 to m.

\( (\frac{1}{n}\sum_{k=1}^{n}\frac{k^3}{n^3}f(k/n))(\frac{1}{m}\sum_{q=1}^{m}f(q/m))+(\frac{1}{n}\sum_{k=1}^{n}f(k/n))(\frac{1}{m}\sum_{q=1}^{m}\frac{q^3}{m^3}f(q/m))- \)
\( -(\frac{1}{n}\sum_{k=1}^{n}\frac{k}{n}f(k/n))(\frac{1}{m}\sum_{q=1}^{m}\frac{q^2}{m^2}f(q/m))-(\frac{1}{n}\sum_{k=1}^{n}\frac{k^2}{n^2}f(k/n))(\frac{1}{n}\sum_{q=1}^{m}\frac{q}{m}f(q/m))\geq 0 \)

Take the limit when n tends to +oo
Take the limit when m tends to +oo

you will have the inequality

[Vlad Matei]

Uneori doar doua Cauchy-uri sunt de ajuns.
Primul \( \displaystyle\left(\int_{0}^{1}f(x)dx\right)\left(\int_{0}^{1}x^{2}f(x)dx\right)\geq
\left(\int_{0}^{1}xf(x)dx\right)^{2} \).
Al doilea \( \displaystyle\left(\int_{0}^{1}x^{3}f(x)dx\right)\left(\int_{0}^{1}xf(x)dx\right)\geq\left(\int_{0}^{1}x^{2}f(x)dx\right)^{2} \)
Le inmultim si am terminat.