For the function $f(x,y,z)= xyz$ how close to the point $(0,0,0)$ should one take the point $(x,y,z)$ in order to make $|f(x,y,z)-f(0,0,0)|<0.008$? 
For the function
  $$f(x,y,z)= xyz$$
   how close to the point $(0,0,0)$ should one take the point $(x,y,z)$ in order to make $|f(x,y,z)-f(0,0,0)|<0.008$?

here is my solution:
after substitution in the absolute value inequality we get
$|xyz|<0.008$ $\Rightarrow$ $\sqrt {x^{2}y^{2}z^{2}}<0.008$ 
but for $x,y,z <1$ 
$\sqrt {x^{2}y^{2}z^{2}} < \sqrt{x^{2}+y^{2}+z^{2}}$
And If we choosed the distance to be less than $\delta = 0.008$ the condition will hold.
the solution of the book is $\delta = 0.2\sqrt{3}$
my questions are :
Is my solution correct? 
how to obtain the solution of the book? 

To get the solution of the book you may use GM-AM:


*

*$\sqrt[3]{abc}\leq \frac{a+b+c}{3}$
$$|xyz| = \sqrt{x^2y^2z^2} \stackrel{GM-AM}{\leq}\sqrt{\frac{(x^2+y^2+z^2)^3}{3^3}} \stackrel{!}{<}0.008 = 0.2^3$$
Now, isolate $\sqrt{x^2+y^2+z^2}$:
$$\sqrt{\frac{(x^2+y^2+z^2)^3}{3^3}} <0.2^3 \Leftrightarrow \sqrt{x^2+y^2+z^2} < \left(0.2^3\sqrt{3^3} \right)^{\frac{1}{3}}=0.2\sqrt{3}$$
So, $\delta =0.2\sqrt{3}$ does the job.
