Question

If three positive real numbers $x,y,z$ satisfy $y-x=z-y$ and $xyz=4$, 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

Answer: (B)

$2^{\frac{2}{3}}$

Since $y-x=z-y$
$\therefore$ $x,y$ and $z$ are in AP.
Let $x,y$ and $z$ are $(a-d),\left(a\right)$ and $(a+d)$.
Again, $xyz=4$
$\Rightarrow (a-d)a(a+d)=4$

$\Rightarrow a(a^2-d^2)=4$

$\Rightarrow a^2-d^2=\frac{4}{a}\Rightarrow d^2=a^2-\frac{4}{a}$
For minimum possible value of $y$, which is $a$, value of the above expression has to be minimum.
As $d^2$ cannot be negative, so minimum possible value of $d^2=0.$
$\therefore d=0$
That is, $a^2-\frac{4}{a}=0$

$\Rightarrow a^2=\frac{4}{a}$

$\Rightarrow a^3=2^2$
$\Rightarrow a=(2^2{)}^{\frac{1}{3}}=2^{\frac{2}{3}}$