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
Inegalitate intre inversul unei sume si al unui produs
Moderators: Bogdan Posa, Laurian Filip
-
Claudiu Mindrila
- Fermat
- Posts: 520
- Joined: Mon Oct 01, 2007 2:25 pm
- Location: Targoviste
- Contact:
Inegalitate intre inversul unei sume si al unui produs
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
-
Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Re: Inegalitate intre inversul unei sume si al unui produs
\( \frac{9}{x+y+z}-\frac{1}{xyz}\le2\ \Longleftrightarrow\ (1+2xyz)(x+y+z)\ge 9xyz\ (*)\ . \)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).
\( \left\|\ \begin{array}{c}
1+2xyz=1+xyz+xyz\ \ge\ 3\cdot\sqrt[3]{1\cdot xyz\cdot xyz}=3\cdot\sqrt[3]{(xyz)^2}\\\\\\\\
x+y+z\ \ge\ 3\cdot\sqrt[3]{(xyz)}\end{array}\ \right\|\ \odot\ \Longrightarrow\ (*)\ . \)
Vezi si aici.