Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1419

\[\boxed{\mathbf{1419}\mathbf{.}}\]

\[1)\ (a + b)(ab + 1) \geq 4ab\]

\[a > 0;\ \ \ b > 0;\]

\[y = ab:\]

\[\left( \frac{y}{b} + b \right)(y + 1) \geq 4y\]

\[\frac{y^{2}}{b} + \frac{y}{b} + by + b - 4y \geq 0\]

\[\frac{y^{2} + y + b^{2}y + b^{2} - 4by}{b} \geq 0\]

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

\[\frac{(y - b)^{2} + y\left( 1 - 2b + b^{2} \right)}{b} \geq 0\]

\[\frac{(y - b)^{2} + y(1 - b)^{2}}{b} \geq 0\]

\[\frac{(ab - b)^{2} + ab(1 - b)^{2}}{b} \geq 0.\]

\[Неравенство\ доказано.\]

\[2)\ a^{4} + 6a^{2}b^{2} + b^{4} > 4ab\left( a^{2} + b^{2} \right);\ \]

\[a \neq b;\]

\[\left( a^{4} + 2a^{2}b^{2} + b^{4} \right) + 4a^{2}b^{2} > 4ab\left( a^{2} + b^{2} \right)\]

\[\left( a^{2} + b^{2} \right)^{2} - 4ab\left( a^{2} + b^{2} \right) + 4a^{2}b^{2} > 0\]

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

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

\[(a - b)^{4} > 0.\]

\[Неравенство\ доказано.\]