Докажите неравенство: (a^2+3)/корень из (a^2+2)>=2.

Ответ:

\[\frac{a^{2} + 3}{\sqrt{a^{2} + 2}} \geq 2\ \]

\[a^{2} + 3 > 0\ \ \ \ \]

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

\[\left( \frac{a^{2} + 3}{\sqrt{a^{2} + 2}} \right)^{2} \geq 4\]

\[\frac{a^{4} + 6a^{2} + 9 - 4a^{2} - 8}{a^{2} + 2} \geq 0\]

\[\frac{a^{4} + 2a^{2} + 1}{a^{2} + 2} \geq 0\]

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

\[\left( a^{2} + 1 \right) > 0;\ \ \]

\[a^{2} + 2 > 0 \Longrightarrow верно.\]