Question

For $x$, $y$, and $z$ positive real numbers, what is the maximum possible value for\[\sqrt{\frac{3x 4y}{6x 5y 4z}} \sqrt{\frac{y 2z}{6x 5y 4z}} \sqrt{\frac{2z 3x}{6x 5y 4z}}

Answer 1

Infinity

Explanation:

Since x,y,z are positive rel numbers, we have that

`(3x4y)/(6x5y4z)=(1)/(10z)\\\\\\(y2z)/(6x5y4z)=(1)/(60x)\\\\\\(2z3x)/(6x5y4z)=(1)/(20y)`

Hence,

`\sqrt{(3x4y)/(6x5y4z)}\sqrt{(y2z)/(6x5y4z)}\sqrt{(2z3x)/(6x5y4z)}\\\\\\=\sqrt{(1)/(10z)(1)/(60x)(1)/(20y)}=\sqrt{(1)/(12000xyz)}`

Now let

`f(x,y,z)=\sqrt{(1)/(12000 xyz)}`

if we take x=y=1, we have

`f(1,1,z)=\sqrt{(1)/(12000z)}`

and so f(1,1,z) tends to infinity as z goes to 0.