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.

Substitute $a=b^2$, then $a^3-a^2-2\sqrt a+2=b^6-b^4-2b+2$, which has $1$ as a root, so we can factor it out:
$$b^6-b^4-2b+2=(b-1)(b^5+b^4-2)$$
The polynomial $b^5+b^4-2$ has still one as a root, so factor more:
$$b^5+b^4-2=(b-1)(b^4+2b^3+2b^2+2b+2)$$
So $a^3-a^2-2=(b-1)^2(b^4+2b^3+2b^2+2b+2)$, which is nonnegative for $b\ge 0$.
