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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.

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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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  • $\begingroup$ Awesome proof, I remember some other problems having similar solutions, they are all beautiful. $\endgroup$ – VividD Jan 20 '15 at 22:50
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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 $b\ge 0$.

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