<s>[INST] 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.
 [/INST] A bit late answer but I thought it might be worth noting.
I was wondering whether we could squeeze out the minimum directly from Young's inequality:
$$ab\leq \frac{a^p}{p} + \frac{b^q}{q};\: a,b \geq 0;\: p,q>1 \text{ and } \frac 1p + \frac 1q = 1$$
And, yes indeed, this works well, too. Since we need the powers of $x$ to cancel out, I start constructing the exponents with $x^6$:
\begin{eqnarray*} x^3 + \frac 1{x^2}
& = & \left(x^6\right)^{\frac 12} + \left(\frac 1{x^6}\right)^{\frac 13}\\
& = & \left(\sqrt[5]{x^6}\right)^{\frac 52} + \left(\frac 1{\sqrt[5]{x^6}}\right)^{\frac 53}\\
& = & \frac 25\cdot \left(\sqrt[5]{\frac{5^2}{2^2}}\sqrt[5]{x^6}\right)^{\frac 52} + \frac 35\cdot\left(\sqrt[5]{\frac{5^3}{3^3}}\frac 1{\sqrt[5]{x^6}}\right)^{\frac 53} \\
& \stackrel{Young}{\geq} & \sqrt[5]{\frac{5^2}{2^2}\cdot \frac{5^3}{3^3}} = \frac 5{\sqrt[5]{2^2\cdot 3^3}}
\end{eqnarray*}
Equality holds for $a^p = b^q$ which means
$$\frac 52 x^3 = \frac 53 \frac 1{x^2} \Leftrightarrow x=\sqrt[5]{\frac 23}$$
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