O inegalitate draguta, cu aplicatii

[Alin Galatan]

In memoria domnului profesor Alexandru Lupas.

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

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
\( \frac{1}{(y+1)^2} \geq \frac{x}{x+y} \cdot \frac{1}{1+xy}. \)
Analog deducem \( \frac{1}{(x+1)^2} \geq \frac{y}{x+y} \cdot
\frac{1}{1+xy}. \) Insumand cele doua inegalitati obtinem
inegalitatea din enunt.

Aplicatii:

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

Solutie Cu substitutiile \( x = \frac{b}{a},
y = \frac{c}{b}, z = \frac{a}{c} \) avem \( xyz=1 \). Astfel
\( (1) \Leftrightarrow
\frac{3+x}{(1+x)^2}+\frac{3+y}{(1+y)^{2}}+\frac{3+z}{(1+z)^{2}} \geq
3 \Leftrightarrow

\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 \)

\( \Leftrightarrow \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}+ \left(
\frac{1}{(x+1)^{2}}+\frac{1}{(y+1)^{2}}\right) + \left(
\frac{1}{(y+1)^{2}}+\frac{1}{(z+1)^{2}}\right) + \left(
\frac{1}{(z+1)^{2}}+\frac{1}{(x+1)^{2}} \right) \geq 3. \)

Folosind lema si tinand cont ca \( \frac{1}{x+1}+\frac{1}{yz+1} = 1 \) si analoagele ei rezulta enuntul.

Probleme propuse:

2. [Vasile Cartoaje] Fie \( a,b,c,d \) numere reale pozitive astfel incat \( abcd=1 \). Sa se arate ca:

\( \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1. \)

Bibliografie:

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

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

[3.] P.K.Hung - Secrets in Inequalities, GIL Publishing House, 2007


P.S. Invit cititorii de pe forum sa completeze aceasta incercare de nota.

[maxim bogdan]

Inegalitatea generala este:

\( \frac{1}{(x+y)^2}+\frac{1}{(x+z)^2}\ge \frac{1}{x^2+yz} \) cu demonstratie analoaga.

[maxim bogdan]

Avem:\( \left (\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\right )+\left (\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}\right )\ge \frac{1}{1+ab}+\frac{1}{1+cd}=\frac{ab+cd+abcd+1}{(ab+1)(cd+1)}=1 \)