### 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.

$\frac{4x^3}{y}+\frac{y+1}{x}$

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

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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.

$\frac{4x^3}{y}+\frac{y+1}{x}$

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

Posted: **Thu Sep 27, 2018 7:31 pm**

Apparently, the denominators should be 1. So the minimum value will be 6.

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.

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.

Posted: **Mon Dec 17, 2018 6:11 pm**

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.

Posted: **Sun Sep 15, 2019 1:49 am**

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".