Page 1 of 1
Inegalitate intre inversul unei sume si al unui produs
Posted: Thu Feb 26, 2009 9:19 am
by Claudiu Mindrila
Sa se demonstreze ca pentru orice numere reale \( x,y,z \) strict pozitive avem inegalitatea: \( \frac{9}{x+y+z}-\frac{1}{xyz}\le2 \).
Cristinel Mortici, G.M.-B. 9/2005
Posted: Thu Feb 26, 2009 10:22 am
by Marius Mainea
Notand \( \sqrt[3]{xyz}=t \) si aplicand AM-GM avem
\( LHS\le \frac{9}{3\sqrt[3]{xyz}}-\frac{1}{xyz}=\frac{3}{t}-\frac{1}{t^3}\le 2 \)
deoarece \( (t-1)^2(2t+1)\ge 0 \)
Re: Inegalitate intre inversul unei sume si al unui produs
Posted: Thu Feb 26, 2009 10:51 am
by Virgil Nicula
Claudiu Mindrila wrote:Sa se demonstreze ca pentru orice numere reale \( x,y,z \) strict pozitive avem
inegalitatea: \( \frac{9}{x+y+z}-\frac{1}{xyz}\le2 \) (Cristinel Mortici, G.M.-B. 9/2005).
\( \frac{9}{x+y+z}-\frac{1}{xyz}\le2\ \Longleftrightarrow\ (1+2xyz)(x+y+z)\ge 9xyz\ (*)\ . \)
\( \left\|\ \begin{array}{c}
1+2xyz=1+xyz+xyz\ \ge\ 3\cdot\sqrt[3]{1\cdot xyz\cdot xyz}=3\cdot\sqrt[3]{(xyz)^2}\\\\\\\\
x+y+z\ \ge\ 3\cdot\sqrt[3]{(xyz)}\end{array}\ \right\|\ \odot\ \Longrightarrow\ (*)\ . \)
Vezi si
aici.