Докажите неравенство (ab+4)(1/a+9/b)>=24; если a>0; b>0.

Ответ:

\[(ab + 4)\left( \frac{1}{a} + \frac{9}{b} \right) \geq 24\]

\[По\ теореме\ Коши:\frac{a + b}{2} \geq \sqrt{\text{ab}}.\]

\[\frac{ab + 4}{2} \geq \sqrt{ab \cdot 4}\]

\[\frac{\frac{1}{a} + \frac{9}{b}}{2} \geq \sqrt{\frac{1}{a} \cdot \frac{9}{b}}\]

\[\frac{ab + 4}{2} \cdot \frac{\frac{1}{a} + \frac{9}{b}}{2} \geq \sqrt{ab \cdot 4} \cdot \sqrt{\frac{1}{a} \cdot \frac{9}{b}}\]

\[(ab + 4)\left( \frac{1}{a} + \frac{9}{b} \right) \geq 4 \cdot \sqrt{\frac{ab \cdot 4 \cdot 9}{\text{ab}}}\]

\[(ab + 4)\left( \frac{1}{a} + \frac{9}{b} \right) \geq 4 \cdot 6\]

\[(ab + 4)\left( \frac{1}{a} + \frac{9}{b} \right) \geq 24\]

\[Что\ и\ требовалось\ доказать.\ \]

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