Inegalitate intre inversul unei sume si al unui produs

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Claudiu Mindrila
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Inegalitate intre inversul unei sume si al unui produs

Post 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
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
Marius Mainea
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Post 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 \( (t-1)^2(2t+1)\ge 0 \)
Virgil Nicula
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Re: Inegalitate intre inversul unei sume si al unui produs

Post 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.
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