Докажите неравенство: (a+3)^2/6≥a-2.

\[\frac{(a + 3)^{2}}{6} \geq a - 2\]

\[(a + 3)^{2} \geq 6a - 12\]

\[a^{2} + 6a + 9 - 6a + 12 \geq 0\]

\[a² + 21 \geq 0.\]