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Lizu12
Pe: 2016-08-12T15:29:07+03:00 In: In:

Inegalitate 2

Buna ziua! M-as bucura daca ati avea rabdare sa va uitati daca este corect ceea ce am rezolvat pana acum si apoi sa.mi spuneti cum as putea continua:

Demonstrati ca : \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\geq\frac{3}{2},unde:a,b,c>0;abc=1

\frac{1}{a^3(b+c)}=\frac{a^2b^2c^2}{a^3(b+c)}=\frac{b^2c^2}{a(b+c)}
\frac{1}{b^3(c+a)}=\frac{a^2b^2c^2}{b^3(a+c)}=\frac{a^2c^2}{b(c+a)}
\frac{1}{c^3(a+b)}=\frac{a^2b^2c^2}{c^3(a+b)}=\frac{a^2b^2} {c(a+b)}

Notez: bc=x_1, ac=x_2, ab=x_3
a(b+c)=z_1,b(c+a)=z_2,c(a+b)=z_3

=> ceea ce trebuie demonstrat este acum:\frac{x_1^2}{z_1}+\frac{x_2^2}{z_2}+\frac{x_3^2}{z_3}\geq \frac{3}{2}

Folosind inegalitatea Cauchy-Buniakovski-Schwarz:
=>
(x_1+x_2+x_3)^2=(\sqrt{z_1}*\frac{x_1}{\sqrt{z_1}}+\sqrt{z_2}*\frac{x_2}{\sqrt{z_2}}+{\sqrt{z_1}}+\sqrt{z_3}*\frac{x_3}{\sqrt{z_3}})^2\leq(z_1+z_2+z_3)(\frac{x_1^2}{z_1}+\frac{x_2^2}{z_2}+\frac{x_3^2}{z_3})
=>\frac{x_1^2}{z_1}+\frac{x_2^2}{z_2}+\frac{x_3^2}{z_3}\geq\frac{(x_1+x_2+x_3)^2}{z_1+z_2+z_3}

deci:\frac{b^2c^2}{a(b+c)}+\frac{a^2c^2}{b(c+a)}+\frac{a^2b^2}{c(a+b)}\geq\frac{(bc+ac+ab)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(bc+ac+ab)^2}{2(bc+ac+ab)}=\frac{bc+ac+ab}{2}

Nu am ajuns la ceea ce era convenabil:\frac{b^2c^2}{a(b+c)}+\frac{a^2c^2}{b(c+a)}+\frac{a^2b^2}{c(a+b)}\geq\frac{abc+abc+abc}{2}=\frac{3}{2}

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3 raspunsuri

  1. gigelmarga
    profesor

    Lizu12 wrote:

    Nu am ajuns la ceea ce era convenabil:\frac{b^2c^2}{a(b+c)}+\frac{a^2c^2}{b(c+a)}+\frac{a^2b^2}{c(a+b)}\geq\frac{abc+abc+abc}{2}=\frac{3}{2}

    E simplu de finalizat.

    \frac{b^2c^2}{a(b+c)}+\frac{a^2c^2}{b(c+a)}+\frac{a^2b^2}{c(a+b)}\geq\frac{ab+bc+ca}{2}\ge \frac{3}{2}\sqrt[3]{a^2b^2c^2}=\frac32.

    BTW, inegalitatea e de la IMO 1995 (Toronto).

  2. Lizu12

    Multumesc !!

  4. gigelmarga
    profesor

    Mai scurt,
    \sum \frac{1}{a^3(b+c)}=\sum\frac{\frac{1}{a^2}}{a(b+c)}\ge \frac{\left(\sum \frac{1}{a}\right)^2}{2\sum ab}=\frac{ab+bc+ca}{2} , etc.

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