<s>[INST] Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a}$ Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a},a \geq 0.$
One  way to do it is using  the formula
$$
a^3+2 - a^2-2 \sqrt{a}=(\sqrt{a}-1)^2(1+(a+1)(\sqrt{a}+1)^2) \geq 0.
$$
But I  hope there is a better way.
 [/INST] You can do it with the AM-GM inequality:
$$
\frac{a^3+a^3+1}{3}\ge \left(a^3\cdot a^3\cdot 1\right)^{\frac{1}{3}}=a^2
$$
$$
2\cdot\frac{a^3+1+1+1+1+1}{6}\ge 2\cdot\left(a^3\cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \right)^{\frac{1}{6}}=2\sqrt{a}
$$
Adding these inequalities yields the desired result.
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