using mean inequality to determine min value of equation

[hearts123]

Hi everyone!
How do I determine the minimum value of y = x^2 + 16/x using the mean inequality?
x is greater than 0. I'm not supposed to do this graphically/with a graphing calculator.
the mean inequality is: (a+b)/2 is greater or equal to sqrt(ab)

I think the first step would be to use x and y as the (a, b) values.
And since x and y aren't equal, it would be:
(x+y)/2 > sqrt(xy)
But I'm not sure how to continue/what to do next.
Any help is greatly appreciated!

 

[Harry_the_cat]

I would let \(\displaystyle a=x^2\) and \(\displaystyle b=\frac{16}{x}\) and sub into the mean inequality \(\displaystyle a+b \geq2\sqrt{ab}\).

Give that a go.

Are you sure the original equation is \(\displaystyle y=x^2 +\frac{16}{x} \) and not \(\displaystyle y=x^2 +\frac{16}{x^2}\)?

 

[hearts123]

I would let \(\displaystyle a=x^2\) and \(\displaystyle b=\frac{16}{x}\) and sub into the mean inequality \(\displaystyle a+b \geq2\sqrt{ab}\).

Give that a go.

Are you sure the original equation is \(\displaystyle y=x^2 +\frac{16}{x} \) and not \(\displaystyle y=x^2 +\frac{16}{x^2}\)?

yeah I'm sure, thanks, I'll give that a go!

 

[pka]

How do I determine the minimum value of y = x^2 + 16/x using the mean inequality?
x is greater than 0. I'm not supposed to do this graphically/with a graphing calculator.
the mean inequality is: (a+b)/2 is greater or equal to sqrt(ab)

Thanks to Harry_the_cat for contributing a suggestion to this thread.
I hope more will offer suggestions. I freely admit that the OP really gives me pause.
I have taught a seminar using An Introduction to Inequalities by Beckenbach&Bellman in which they give a great attention to this topic( it is even on the cover of the book). However, I have never seen it applied to a minimum question.
So others please post ideas.

 

[MarkFL]

I'm wondering if we actually have:

[MATH]y=\frac{x^2+16}{x}[/MATH]

 

[Harry_the_cat]

MarkFL, the poster appeared to confirm the function after my post. But yes, the method I suggested would work for that interpretation.