Question

Given that $x,y,z$ are positive real such that $xyz=32$. If the minimum value of $x^2+4xy+4y^2+2z^2$ is equal to $m$, then the value of $\frac{m}{16}$ is?

Answer 1

Find the value $m$ to get $\frac{m}{16}$:

Now, $x^2+4xy+4y^2+2z^2=x^2+2xy+2xy+4y^2+z^2+z^2$

Since Arithmetic Mean $\ge$ Geometric Mean

$$\begin{gathered} \therefore \frac{x^2+2xy+2xy+4y^2+z^2+z^2}{6}\ge {\left(16x^4{y}^4{z}^4\right)}^{\frac{1}{6}} \\ {x}^2+2xy+2xy+4y^2+z^2+z^2\ge 6{(2^4\times {32}^4)}^{\frac{1}{6}} \\ {x}^2+2xy+2xy+4y^2+z^2+z^2\ge 6{(2^4\times {\left(2^5\right)}^4)}^{\frac{1}{6}} \\ {x}^2+2xy+2xy+4y^2+z^2+z^2\ge 6{(2^4\times 2^{20})}^{\frac{1}{6}} \\ {x}^2+2xy+2xy+4y^2+z^2+z^2\ge 6{(2^{24})}^{\frac{1}{6}} \\ {x}^2+2xy+2xy+4y^2+z^2+z^2\ge 6\times 16 \end{gathered}$$

Thus $m=6\times 16$,

So, $\frac{m}{16}=6$

Hence, the value of $\frac{m}{16}$ for $x^2+4xy+4y^2+2{z^2}^{}$is $6$.