Inegalitate

[Claudiu Mindrila]

Sa se demonstreze ca: \( \left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}b}+\frac{1}{b^{2}c}+\frac{1}{c^{2}a}\right)\ge27 \) oricare ar fi numerele reale pozitive \( a,\ b, \ c \).

Stefan Smarandache, concursul revistei "Arhimede", etapa finala, 2009

[Marius Dragoi]

Avem: \( \sum_{} {a^3} + \sum_{} {a^2c} \geq 2 (\sum_{}{a^2b}) \)
Atunci inegalitatea de demonstrat devine: \( 3(\sum_{}{a^2b})(\sum_{}{\frac {1}{a^2b}}) \geq 27 \) \( \Leftrightarrow \) \( (\sum_{}{a^2b})(\sum_{}{\frac {1}{a^2b}}) \geq 9 \)
Dar \( \sum_{}{\frac {1}{a^2b} \geq \frac {9}{\sum_{}{a^2b} \) de unde rezulta concluzia.

[Claudiu Mindrila]

Solutia 1. Cu inegalitatea Holder avem: \( \left(\sum_{cyc}a\right)\left(\sum_{cyc}a^{2}\right)\left(\sum_{cyc}\frac{1}{a^{2}b}\right)=\left(\sum_{cyc}\left(\sqrt[3]{a}\right)^{3}\right)\left(\sum_{cyc}\left(\sqrt[3]{a^{2}}\right)^{3}\right)\left(\sum_{cyc}\left(\sqrt[3]{\frac{1}{a^{2}b}}\right)^{3}\right)\ge\left(\sum_{cyc}\sqrt[3]{a\cdot a^{2}\cdot\frac{1}{a^{2}b}}\right)^{3}=\left(\sum_{cyc}\sqrt[3]{\frac{a}{b}}\right)^{3} \), iar apoi cu inegalitatea mediilor \( \left(\sum_{cyc}\sqrt[3]{\frac{a}{b}}\right)^{3}\ge\left(3\cdot\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}\right)^{3}=27 \).

Solutia 2. Cu inegalitatea mediilor avem: \( \left|\begin{array}{c}
a+b+c\ge3\sqrt[3]{abc}\\
a^{2}+b^{2}+c^{2}\ge3\sqrt[3]{a^{2}b^{2}c^{2}}\\
\frac{1}{a^{2}b}+\frac{1}{b^{2}c}+\frac{1}{c^{2}a}\ge3\sqrt[3]{\frac{1}{a^{2}b}\cdot\frac{1}{b^{2}c}\cdot\frac{1}{c^{2}a}}=3\sqrt[3]{\frac{1}{a^{3}b^{3}c^{3}}}\end{array}\right|\odot\Longrightarrow\left(\sum_{cyc}a\right)\left(\sum_{cyc}a^{2}\right)\left(\sum_{cyc}\frac{1}{a^{2}b}\right)\ge27. \)