How to solve this system of inequalities and equalities systematically:
$\begin{array}{l} \left\{ {\begin{array}{*{20}{c}} {\frac{x}{{2x + 3y}} + \frac{y}{{y + z}} + \frac{z}{{x + z}} = \frac{{34}}{{33}}}\\ {1 \le x \le 4}\\ {1 \le y \le 4}\\ {1 \le z \le 4}\\ {x \ge y}\\ {x \ge z} \end{array}} \right.\\ \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {x = 4}\\ {y = 1}\\ {z = 2} \end{array}} \right. \end{array}$
I got this answer from Mathematica, but I am not so sure how to do it manually.
Please help me with this, thank you for your enthusiasm !
Multiply by $(2x+3y)(y+z)(x+z)$ to clear fractions, collect terms, and thereby conclude that:
$$\frac{x}{2x + 3y} + \frac{y}{y + z} + \frac{z}{x + z} - \frac{34}{33} = \frac{31 x^2 y - 35 x^2 z - 3 x y^2 - 5 x y z + 31 x z^2 + 96 y^2 z - 3 y z^2}{33 (2 x + 3 y) (x + z) (y + z)}.$$
It is easy to see that $(2 x + 3 y) (x + z) (y + z) > 0$ for your range of $x,y,z$.
Therefore, your problem amounts to finding $x,y,z$ that satisfy $f(x,y,z)=0$, where
$$f(x,y,z) = 31 x^2 y - 35 x^2 z - 3 x y^2 - 5 x y z + 31 x z^2 + 96 y^2 z - 3 y z^2,$$
subject to $1 \le y \le x \le 4$, $1 \le z \le x \le 4$.
Writing $f(x,y,z) = p(x,y) + q(x,y) z + r(x,y) z^2$ for appropriate polynomials $p(x,y),q(x,y),r(x,y)$, we can then apply the quadratic formula (w.r.t. $z$) to obtain
$$z = \frac{35 x^2 + 5 x y - 96 y^2 \pm \sqrt{s(x,y)}}{62 x - 6 y},$$
where
$$s(x,y) = 1225 x^4 - 3494 x^3 y - 5951 x^2 y^2 - 996 x y^3 + 9216 y^4.$$
Equivalently, $s(x,y) = y^4 \cdot t(x/y)$ where
$$t(w) = 1225 w^4 - 3494 w^3 - 5951 w^2 - 996 w + 9216.$$
We have $t(w) = (w-4)(w-1)(1225w^2 + 2631w + 2304)$, so the only real roots of $t(w)$ are at $w=1$ and $w=4$. We have $t(w)<0$ for $1<w<4$, so this region will not lead to any valid solutions for the original problem, as then the formula for $z$ above gives imaginary values for $z$, which are not allowable. Therefore, we must have either $w=1$ or $w=4$, which corresponds to solutions where $x/y=1$ or $x/y=4$. We can consider each of those two cases separately:
If $x/y=1$: Plugging in $y=x$ into the formula above, we discover
$$z = \frac{35 x^2 + 5 x^2 - 96 x^2}{62 x - 6 x} = -x.$$
Therefore, this case is impossible, as it is inconsistent with the requirement that $z \ge 1$.
If $x/y=4$: This is only possible if $x=4$, $y=1$. Plugging into the formula for $z$ above, we find $z=2$. This is the solution in your question.