Question

The number of different possible values for the sum $x+y+z$, where $x,y,z$ are real numbers such that $x^4+4y^4+16z^4+64=32xyz$ is

• A. $1$
• B. $2$
• C. $4$
• D. $8$

Answer 1

The correct option is C $4$

$\begin{array}{c}S=x+y+z=?\\ x^4+4y^4+16z^4+64=32xyz\\ \frac{x^4+4y^4+16z^4+64}{4}\ge {\left(x^4\cdot 4y^4\cdot 16z^4\cdot 64\right)}^{\frac{1}{4}}\\ \frac{32xyz}{4}\ge 8xyz\\ 8xyz\ge 8xyz\\ AM=GM\\ \therefore x^4=4y^4=16z^4=64\\ x=\pm 2\sqrt{2}\\ y=\pm 2\\ z=\pm \sqrt{2}\end{array}$

But $32xyz$ is positive. Hence, number of possible sums are $4$.