CAT 2003 | Question: 2-89

Question

If three positive real numbers $x, y, z$ satisfy $y – x = z – y$ and $x y z = 4,$ then what is the minimum possible value of $y?$

• A. $2^{\frac{1}{3}}$
• B. $2^{\frac{2}{3}}$
• C. $2^{\frac{1}{4}}$
• D. $2^{\frac{3}{4}}$

Answer 1

y - x = z - y

x + z = 2y$\dots(1)$

xyz =4 $\dots(2)$

 
We know that, AM  >= GM

Hence, $\frac{x+y+z}{3}$  >= $(xyz)^{\frac{1}{3}}$

From (1) and (2)

$\frac{3y}{3}$ >= $4^{\frac{1}{3}}$

y  >= $2^{\frac{2}{3}}$

 

Hence,Option(B)$2^{\frac{2}{3}}$ is the correct choice.