This is related to Trig?

Using a bit of algebraic manipulation, we can rewrite the equation as follows: $(x+y+z - 2)^2 = (2-x)(2-y)(2-z)$ Now since the left side of the equation is nonnegative, so is the right side. Hence we can apply the AM-GM inequality to get: $(x+y+z - 2)^2 = (2-x)(2-y)(2-z) \le \left(\frac{6 - (x+y+z)}{3}\right)^3$ Thus if we let $s = x+y+z$ , we get the following inequality: $(s-2)^2 \le \frac{(6-s)^3}{27}$ $27(s-2)^2 + (s-6)^3 \le 0$ $s^3+9s^2-108 \le 0$ $(s-3)(s+6)^2 \le 0$

Therefore $s \le 3$ . Therefore the maximum value of $x+y+z$ is $\boxed{3}$ . Note that this value is achieved when $x = y = z = 1$ .

It can be shown that $(x, y, z) = (2\cos A, 2\cos B, 2\cos C),$ where $A + B + C = \pi$ and $A, B, C < \frac{\pi}{2}$ satisfy the condition given in the problem. Thus, $x + y + z = 2(\cos A + \cos B + \cos C) \leq 2\left(\frac {3}{2}\right) = \boxed{3}.$

. very simple the equation equates to 4 therefore the answer should not exceed or be a little less than 4. since there are 4 terms then it can be 1 multiplied by 4. conclusion, the variables are equal and their value is 1.

(If you didn't manage to see the trig solution like me).

Trying to get the equation into an easier one with only 2 variables, we substitute z=2. Note that we cannot substitute z=-2 since it has to be positive.

Now, that gives us $(x+y)^2=0$ or $x+y=0$ .

Our sum in that case would be 2.

We cannot substitute a value for z greater than 2 (otherwise either x or y must be negative which violates the conditions).

Therefore, we try z=1. This gives us $x^2+xy+y^2=3$ .

$y \ge 2$ would give us negative values for x, so y=1. Following from this, x=1.

Our sum here is 3, which is greater than the other case of 2.

Therefore, the maximum value of $x+y+z$ is $\boxed{3}$ .