Question

Given that $x,y,z$ are positive real numbers such that $xyz=32$. The minimum value of $x^2+4xy+4y^2+2z^2$ is equal to

• A. $64$
• B. $256$
• C. $96$
• D. $216$

Answer 1

The correct option is A $64$

Since $AM\ge GM$
$\frac{x^2+2xy+2xy+4y^2+z^2+z^2}{6}$ $\ge {(4^2.x^4.y^4.z^4)}^{\frac{1}{6}}$
$\therefore x^2+2xy+2xy+4y^2+z^2+z^2\ge {6.2}^{4/6}.(xyz{)}^{2/3}$
$x^2+4xy+4y^2+2z^2\ge 96$