ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 1025

\[\boxed{\text{1025.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[a > 0,\ \ b > 0\]

\[\textbf{а)}\ (a + b)\left( \frac{1}{a} + \frac{1}{b} \right) \geq 4\]

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

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

\[a^{2} + 2ab + b^{2} \geq 4ab\]

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

\[(a - b)^{2} \geq 0 \Longrightarrow верно,\ ч.т.д.\]

\[\textbf{б)}\frac{a}{b^{2}} + \frac{b}{a^{2}} \geq \frac{1}{a} + \frac{1}{b}\]

\[\frac{a^{3} + b^{3}}{a^{2}b^{2}} \geq \frac{a + b}{\text{ab}}\]

\[\frac{a^{3} + b^{3}}{a^{2}b^{2}} - \frac{a + b}{\text{ab}} \geq 0\]

\[\frac{(a + b)\left( a^{2} - ab + b^{2} - ab \right)}{a^{2}b^{2}} \geq 0\]

\[\frac{(a + b)\left( a^{2} - 2ab + b^{2} \right)}{a^{2}b^{2}} \geq 0\]

\[\frac{(a + b)(a - b)^{2}}{a^{2}b^{2}} \geq 0 \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (a - b)^{2} \geq 0 - верно \\ a^{2}b^{2} > 0 - верно\ \ \ \ \ \ \ \\ a + b \geq 0 - верно\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[a + b \geq 0;\ \ так\ как\ a > 0\ и\ \]

\[b > 0 \Longrightarrow ч.т.д.\]