O inegalitate draguta, cu aplicatii
Posted: Thu Sep 27, 2007 10:20 pm
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.
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.