Find minimum of $2x^3+x+\frac3{x^2}$ for $x>0$

Question

Using AM-GM inequality we have: $$2x^3+x+\frac3{x^2}=2x^3+x+\frac3{2x^2}+\frac3{2x^2}\geq 4\sqrt[4]{2x^3x\frac3{2x^2}\frac3{2x^2}}=4\sqrt[4]{\frac92}$$

As it turns out it is not the minimum, can you spot the mistake?

Answer 1

The issue is it assumes both $2x^3=x$ and $x={3\over 2x^3}$ which cannot be both true simultaneously.

Answer 2

There is no flaw in your work - it is true that $$2x^3 + x + \frac{3}{x^2} \ge 4 \sqrt[4]{\frac{9}{2}}$$ for $x > 0$. However, AM-GM inequality won't be enough to find the minimum, as it only attains equality for $$2x^3 = x = \frac{3}{2x^2} = \frac{3}{2x^2}$$

which never happens.

Answer 3

By using AM-GM inequality we get:

$2x^3+x+\dfrac3{x^2}=x^3+x^3+x+\dfrac1{x^2}+\dfrac1{x^2}+\dfrac1{x^2}\geqslant$

$\geqslant x+5\sqrt[5]{x^3\cdot x^3\cdot\dfrac1{x^2}\cdot\dfrac1{x^2}\cdot\dfrac1{x^2}}=x+5\;.$

By using again AM-GM inequality we get:

$2x^3+x+\dfrac3{x^2}=2x^3+\dfrac{16}9x+\dfrac3{x^2}-\dfrac79x=$

$=x^3+x^3+\dfrac89x+\dfrac89x+\dfrac3{4x^2}+\dfrac3{4x^2}+\dfrac3{4x^2}+\dfrac3{4x^2}-\dfrac79x\geqslant$

$\geqslant8\sqrt[8]{x^3\cdot x^3\cdot\dfrac89x\cdot\dfrac89x\cdot\dfrac3{4x^2}\cdot\dfrac3{4x^2}\cdot\dfrac3{4x^2}\cdot\dfrac3{4x^2}}-\dfrac79x=$

$=8\sqrt[8]{\dfrac14}-\dfrac79x=8\sqrt[4]{\dfrac12}-\dfrac79x>6+\dfrac{163}{225}-\dfrac79x\;.$

Consequently,

$2x^3+x+\dfrac3{x^2}\geqslant x+5\geqslant5+\dfrac{97}{100}\quad$ for all $\;x\geqslant\dfrac{97}{100}\;,$

$\begin{align}2x^3+x+\dfrac3{x^2}&>6+\dfrac{163}{225}-\dfrac79x>6+\dfrac{163}{225}-\dfrac79\cdot\dfrac{97}{100}=\\&=6+\dfrac{163}{225}-\dfrac{679}{900}=6+\dfrac{652-679}{900}=\\&=6-\dfrac{27}{900}=6-\dfrac3{100}=5+1-\dfrac3{100}=\\&=5+\dfrac{97}{100}\quad\text{ for all }\;0<x<\dfrac{97}{100}\;.\end{align}$

Hence,

$2x^3+x+\dfrac3{x^2}\geqslant5+\dfrac{97}{100}\quad$ for all $\;x>0\;.$

The value $\;5+\dfrac{97}{100}=5.97\;$ obtained by applying twice the AM-GM inequality, is very close to the minimum of the function $\;f(x)=2x^3+x+\dfrac3{x^2}\;$ on the interval $\;\bigl]0,+\infty\bigr[\;.$