Докажите, что: (a+b)(1/a+1/b)>=4, если a>0, b>0.

Ответ:

\[(a + b)\left( \frac{1}{a} + \frac{1}{b} \right) \geq 4;\ \ \ \ \ \ a > 0,\ \ \ \ \ \]

\[b > 0\]

\[\frac{(a + b)(b + a)}{\text{ab}} \geq 4\]

\[\frac{(a + b)^{2}}{\text{ab}} \geq 4\]

\[\frac{a^{2} + b^{2} + 2ab - 4ab}{\text{ab}} \geq 0\]

\[\frac{(a - b)^{2}}{\text{ab}} \geq 0 \Longrightarrow верно.\]

\[ab > 0,\ т.к.\ a > 0,\ \ \ \ b > 0;\]

\[(a - b)^{2} > 0 \Longrightarrow всегда.\ \]