O inegalitate utila, cu aplicatii

mircea.lascu
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Membru din: Lun Iul 12, 2010 9:02 pm

O inegalitate utila, cu aplicatii

Mesaj de mircea.lascu »

In memoria domnului profesor Alexandru Lupas.

Fie $x,y$ numere pozitive, atunci: $\displaystyle\frac{1}{(1+x)^{2}}+\frac{1}{(1+y)^{2}}\geq\frac{1}{1+xy}.$

Solutie: [D. Grinberg] Din inegalitatea lui Cauchy-Schwarz avem:
$(x+y)(1+xy)\geq (\sqrt{x}+\sqrt{x}y)^{2}=x(1+y)^{2}$ de unde
$\displaystyle\frac{1}{(y+1)^{2}}\geq\frac{x}{x+y}\cdot\frac{1}{1+xy}.$
Analog deducem $\displaystyle\frac{1}{(x+1)^{2}}\geq \frac{y}{x+y}\cdot\frac{1}{1+xy}.$ Insumand cele doua inegalitati obtinem inegalitatea din enunt.

Aplicatii:

1. [Vasile Cirtoaje] Fie $a,b,c$ numere strict pozitive. Sa se arate ca:
$\displaystyle\frac{a(3a+b)}{(a+b)^{2}}+\frac{b(3b+c)}{(b+c)^{2}}+\frac{c(3c+a)}{(c+a)^{2}}\geq 3.\ \ \ (1)$

Solutie: Cu substitutiile $x=\displaystyle\frac{a}{b},y=\frac{c}{b},z=\frac{a}{c}$ avem $xyz=1.$ Astfel
$(1)\Leftrightarrow\displaystyle\frac{3+x}{(1+x)^{2}+\frac{3+y}{(1+y)^{2}}+\frac{3+z}{(1+z)^{2}}\geq$$3\Leftrightarrow\displaystyle\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+\frac{2}{(1+x)^{2}}+\frac{2}{(1+y)^{2}}+\frac{2}{(1+z)^{2}}\geq 3\Leftrightarrow$
$\Leftrightarrow\displaystyle\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+\left(\frac{1}{(x+1)^{2}}+\frac{1}{(y+1)^{2}}\right)+$$\displaystyle\left(\frac{1}{(y+1)^{2}}+\frac{1}{(z+1)^{2}}\right)+\left(\frac{1}{(z+1)^{2}}+\frac{1}{(x+1)^{2}}\right)\geq 3.$
Folosind lema si tinand cont ca $\displaystyle\frac{1}{x+1}+\frac{1}{yz+1}=1$ si analoagele ei rezulta enuntul.

Probleme propuse:

2. [Vasile Cirtoaje] Fie $a,b,c,d$ numere reale pozitive astfel incat $abcd=1.$ Sa se arate ca:
$\displaystyle\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}}\geq 1.$

3. Fie $a,b,c>0$ astfel incat $abc=1$. Aratati ca:
$\displaystyle\frac{1}{(a+1)^{2}}+\frac{1}{(b+1)^{2}}+\frac{1}{(c+1)^{2}}\geq\frac{3}{4}.$

Bibliografie:

[1.] T. Andreescu, V. Cirtoaje, G. Dospinescu si M. Lascu - Old and New Inequalities, GIL Publishing House, 2004

[2.] V. Cirtoaje - Algebraic Inequalities. Old and New Methods, GIL Publishing House, 2006

[3.] P. K. Hung - Secrets in Inequalities. Basic inequalities, GIL Publishing House, 2007
Stefan Tudose
Mesaje: 258
Membru din: Mar Aug 30, 2011 7:25 pm

Re: O inegalitate utila, cu aplicatii

Mesaj de Stefan Tudose »

Problema 2
Folosindu-ne de faptul că $\displaystyle\frac{1}{(1+x)^{2}}+\frac{1}{(1+y)^{2}}\geq\frac{1}{1+xy}$ , oricare ar fi x şi y pozitive,
$\displaystyle\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}\geq\frac{1}{1+ab}$
$\displaystyle\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}}\geq\frac{1}{1+cd}$

Însumându-le,

$\displaystyle\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}}\geq\frac{1}{1+ab}+\frac{1}{1+cd}$
Dar cum $\displaystyle\frac{1}{1+ab}+\frac{1}{1+cd}=\frac{cd}{cd+abcd}+\frac{1}{1+cd}=\frac{cd+1}{1+cd}=1$

inegalitatea este demonstrată.

Egalitatea se obţine pentru a=b=c=d=1.
Ultima oară modificat Dum Feb 26, 2012 2:20 pm de către Stefan Tudose, modificat de 2 ori în total.
Stefan Tudose
Mesaje: 258
Membru din: Mar Aug 30, 2011 7:25 pm

Re: O inegalitate utila, cu aplicatii

Mesaj de Stefan Tudose »

Problema 2

$\left.\begin{array}{l}\displaystyle\frac{1}{(a+1)^2}+\frac{1}{(b+1)^2} \ge \frac{1}{1+ab} \\ \displaystyle\frac{1}{1+ab} = \frac{c}{c+1} \end{array}\right\}\Rightarrow \displaystyle\frac{1}{(a+1)^2}+\frac{1}{(b+1)^2}+\frac{1}{(c+1)^2}$ $\ge \displaystyle \frac{c}{c+1}+\frac{1}{(c+1)^2}$

Inegalitatea de demonstrat devine:

$\displaystyle \frac{c}{c+1}+\frac{1}{(c+1)^2} \ge \frac{3}{4}\Leftrightarrow$

$\Leftrightarrow \displaystyle \frac{c^2+c+1}{(c+1)^2} \ge \frac{3}{4}$

$\Leftrightarrow 4c^2+4c+4 \ge 3c^2+6c+3$

$\Leftrightarrow c^2-2c+1 \ge 0$

$\Leftrightarrow (c-1)^2 \ge 0$ , care este evident adevarat.
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