# Using AM-GM Inequality (Arithmetic mean- Geometric Mean) Find the minimum value of $2x^2+\frac{1}{x^4}$

Problem: The problem is to find the minimum value of $2x^2+\frac{1}{x^4}$ and though you can use any method in finding the answer, you probably want to use AM-GM method

My thought process before getting stuck In this case I know that according to the AM-GM theorem/equation $\frac{a_1+a_2+...a_n}{n}\ge {\root n\of {a_1*a_2....a_n}}$. But then i do not now how to use the AM GM method in finding the minimum value

• Hint: $2x^2+\frac{1}{x^4} = x^2+x^2+\frac{1}{x^4}$. Jul 7 '16 at 7:36

$2x^2+\frac{1}{x^4}=x^2+x^2+\frac{1}{x^4} \geq 3 \sqrt[3]{(x^2)(x^2)\left(\frac{1}{x^4}\right)}=3$
The equation $$\frac{a+b+c}2\left[(b-c)^2+(c-a)^2+(a-b)^2\right]=a^3+b^3+c^3-3abc$$ shows that equality in the AGM $$\sqrt[\large3]{xyz}\le\frac{x+y+z}3$$ occurs only when $x=y=z$.
This in conjunction with Jason M's observation that the AGM applies to this question as $$3\le x^2+x^2+\frac1{x^4}$$ shows that the minimum is $3$ and it is attained when $x=1$.