Докажите неравенство: (a^2+5)/корень из (a^2+4)>=2.

Ответ:

\[\frac{a^{2} + 5}{\sqrt{a^{2} + 4}} \geq 2,\ \ a^{2} + 5 > 0,\ \ \ \]

\[\sqrt{a^{2} + 4} > 0\]

\[\left( \frac{a^{2} + 5}{\sqrt{a^{2} + 4}} \right)^{2} \geq 4\]

\[\frac{a^{4} + 10a² + 25}{a² + 4} \geq 4\]

\[\frac{a^{4} + 10a^{2} + 25 - 4a^{2} - 16}{a^{2} + 4} \geq 0\]

\[\frac{\left( a^{2} + 3 \right)^{2}}{a^{2} + 4} \geq 0 \Longrightarrow верно.\]