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

  1. $2^{(1/3)}$
  2. $2^{(2/3)}$
  3. $2^{(1/4)}$
  4. $2^{(3/4)}$
in Quantitative Aptitude retagged by
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Given that, $x, y, z \in \mathbb{R}^{+}$

  • $y – z = z- y \quad \longrightarrow (1)$
  • $xyz = 4 \quad \longrightarrow (2)$

From equation $(1),$

$\Rightarrow y – x = z – y$

$\Rightarrow \boxed{2y = z + x}$

So, $x,y,z$ are in arithmetic progression.

We know that, $\text{AM} \geq \text{GM}$

$\Rightarrow \frac{x+y+z}{3} \geq \sqrt[3]{xyz}$

$\Rightarrow \frac{y+2y}{3} \geq (4)^{\frac{1}{3}}$

$\Rightarrow \frac{3y}{3} \geq (4)^{\frac{1}{3}}$

$\Rightarrow y \geq (2^{2})^{\frac{1}{3}}$

$\Rightarrow \boxed{y \geq (2)^{\frac{2}{3}}}$

Correct Answer $: \text{B}$

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