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

Answer:

Infinity

Step-by-step explanation:

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

$\dfrac{3x4y}{6x5y4z}=\dfrac{1}{10z}\\\\\\\dfrac{y2z}{6x5y4z}=\dfrac{1}{60x}\\\\\\\dfrac{2z3x}{6x5y4z}=\dfrac{1}{20y}$

Hence,

$\sqrt{\dfrac{3x4y}{6x5y4z}}\sqrt{\dfrac{y2z}{6x5y4z}}\sqrt{\dfrac{2z3x}{6x5y4z}}\\\\\\=\sqrt{\dfrac{1}{10z}\dfrac{1}{60x}\dfrac{1}{20y}}=\sqrt{\dfrac{1}{12000xyz}}$

Now let

$f(x,y,z)=\sqrt{\dfrac{1}{12000 xyz}}$

if we take x=y=1, we have

$f(1,1,z)=\sqrt{\dfrac{1}{12000z}}$

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