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### Minimum value

Posted: Sat Jul 22, 2017 3:18 pm
Determine the minimum value of
$\frac{4x^3}{y}+\frac{y+1}{x}$

with $x>0$ and $y>0$ real numbers.

### Re: Minimum value

Posted: Thu Sep 27, 2018 7:31 pm
Apparently, the denominators should be 1. So the minimum value will be 6.

### Re: Minimum value

Posted: Sat Dec 15, 2018 7:39 pm
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.

### Re: Minimum value

Posted: Mon Dec 17, 2018 6:11 pm
Ragib Farhat Hasan wrote:
Thu Sep 27, 2018 7:31 pm
Apparently, the denominators should be 1. So the minimum value will be 6.
I don't think so. Bcoz, x>0 and y>0 are real numbers not natural.

### Re: Minimum value

Posted: Sun Sep 15, 2019 1:49 am
NABILA wrote:
Mon Dec 17, 2018 6:11 pm
Ragib Farhat Hasan wrote:
Thu Sep 27, 2018 7:31 pm
Apparently, the denominators should be 1. So the minimum value will be 6.
I don't think so. Bcoz, x>0 and y>0 are real numbers not natural.
I know the difference.

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".