Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$

Question

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.

Answer 1

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.

Answer 2

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 $$\frac 52 x^3 = \frac 53 \frac 1{x^2} \Leftrightarrow x=\sqrt[5]{\frac 23}$$

Answer 3

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 $a=\sqrt[5]{2/3}$.