Determine the minimum value of
$\frac{4x^3}{y}+\frac{y+1}{x}$
with $x>0$ and $y>0$ real numbers.
Minimum value

 Posts: 62
 Joined: Sun Mar 30, 2014 10:40 pm
Re: Minimum value
Apparently, the denominators should be 1. So the minimum value will be 6.
Re: Minimum value
No.
If, X>0, y>0
Then, it could be x=y=0.1 or 0.01 or 0.000001
Because, x and y are real number.
If, X>0, y>0
Then, it could be x=y=0.1 or 0.01 or 0.000001
Because, x and y are real number.
Re: Minimum value
I don't think so. Bcoz, x>0 and y>0 are real numbers not natural.Ragib Farhat Hasan wrote: ↑Thu Sep 27, 2018 7:31 pmApparently, the denominators should be 1. So the minimum value will be 6.

 Posts: 62
 Joined: Sun Mar 30, 2014 10:40 pm
Re: Minimum value
I know the difference.NABILA wrote: ↑Mon Dec 17, 2018 6:11 pmI don't think so. Bcoz, x>0 and y>0 are real numbers not natural.Ragib Farhat Hasan wrote: ↑Thu Sep 27, 2018 7:31 pmApparently, the denominators should be 1. So the minimum value will be 6.
What I meant was: in the solution to the problem, both $x$ and $y$ should be 1.
BTW, this is just an observation since I didn't go through with the solution; it is an "apparent result".