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

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

$2x^2+\frac{1}{x^4}=x^2+x^2+\frac{1}{x^4} \geq 3 \sqrt[3]{(x^2)(x^2)\left(\frac{1}{x^4}\right)}=3$

  • $\begingroup$ Good job so far. And to conclude that this is the actual minimum ... (I'm sure you know how to do that. If you decided to leave that step to the OP as an exercise I approve!) $\endgroup$ Jul 7 '16 at 7:47
  • $\begingroup$ @JyrkiLahtonen Yes of course, unless the OP asks, but I think they got it. $\endgroup$
    – JasonM
    Jul 7 '16 at 7:49

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 $x=1$.


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