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.
 A: Very similar to what you have done:
$$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}}{5}\geq
 \sqrt[5]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[5]{\frac{1}{108}}$$
This gives us
$$x^3+\frac{1}{x^2}\geq \sqrt[5]{\frac{1}{108}}$$
Desmos screenshot:

P.S. Your method fails because equality holds only when
$$\frac{x^3}{2}=\frac{1}{3x^2}=1$$ which is impossible. The extra "one" you added in your AM-GM application screwed your attempt.
A: 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
$$\frac 52 x^3 = \frac 53 \frac 1{x^2} \Leftrightarrow x=\sqrt[5]{\frac 23}$$
A: 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 $a=\sqrt[5]{2/3}$.
