Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$ Finding this minimum must be only done using ineaqualities.

$x^3+\frac{1}{x^2}=\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}$
Using inequalities of arithemtic and geometric means:
$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}+1}{6}\geqslant
 \sqrt[6]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[6]{\frac{1}{108}}\Rightarrow x^3+\frac{1}{x^2}\geqslant 6\sqrt[6]{\frac{1}{108}}-1 $

Sadly $\ 6\sqrt[6]{\frac{1}{108}}-1$ is not correct answer, it is not the minimum.

We need to find a positive number $a$ such that
$$x^3+{1\over x^2}\ge a^3+{1\over a^2}$$
for all $x\gt0$. The existence of such an $a$ is not in doubt, since $x^3+{1\over x^2}\to\infty$ as $x\to0$ and $x\to\infty$.  But
$$\begin{align}
x^3+{1\over x^2}\ge a^3+{1\over a^2}
&\iff(x^3-a^3)-\left({1\over a^2}-{1\over x^2}\right)\ge0\\
&\iff(x-a)\left((x^2+ax+a^2)-{x+a\over a^2x^2} \right)\ge0
\end{align}$$
Now $x-a$ changes sign at $x=a$, so in order for nonnegativity to be maintained, the other factor, $(x^2+ax+a^2)-{x+a\over a^2x^2}$, must do so as well. In particular that factor must also equal $0$ at $x=a$, so we must have
$$3a^2={a+a\over a^2a^2}={2\over a^3}$$
or $a=\sqrt[5]{2/3}$.
