# Is there a systematic way to derive constraints in a system of equations?

I have this system of equations:

$$x_1 = y_1+ y_2+ y_3$$

$$x_2 = y_1- y_2$$

$$x_3 = y_1 + y_2 - 2y_3$$

And I have these constraints:

$$y_1 \geq y_2 \geq 0 \geq y_3$$

From the constraints I can derive these constraints for the $$x$$'s:

$$x_2 \geq 0$$

$$x_3 \geq x_1$$

$$x_3 \geq x_2$$

I "derived" these constraints by eye, but they all seem very natural. However, when I put the constraints into Mathematica, I get a different set of constraints:

$$x_2 \geq 0$$

$$x_3 \geq x_1$$

$$2x_1 + x_3 \geq 3x_2$$

It's straightforward to show that Mathematica's 3rd constraint is equivalent to $$y_1 + y_2 \geq y_1 - y_2$$, which in turn is equivalent to $$y_2 \geq 0$$, so it's a valid constraint. However, I don't see why these constraints are different from the ones I derive (they're definitely different - I got Mathematica to integrate an equation subject to these constraints, and the two constraints yield a different answer). I can't tell which constraint is incorrect or why, so I'm asking if there's a systematic way to derive them.

Your answer is not true. Indeed, \begin{align} \left\{\begin{array}{l} x_2 \ge 0\\ x_3 \ge x_1\\ x_3 \ge x_2 \\ \end{array} \right. \Leftrightarrow \left\{\begin{array}{l} y_1-y_2 \ge 0\\ y_1+y_2 - 2y_3 \ge y_1+y_2 + y_3\\ y_1+y_2 - 2y_3 \ge y_1 - y_2 \\ \end{array} \right. \Leftrightarrow \left\{\begin{array}{l} y_1 \ge y_2\\ 0 \ge y_3\\ y_2 \ge y_3 \\ \end{array} \right. \Leftrightarrow (y_1\ge y_2\ge y_3) \land (0 \ge y_3) \end{align} which is different from $$y_1\ge y_2\ge 0\ge y_3$$.
You may get the correct answer as follows. \begin{align} \left\{\begin{array}{l} x_1 = y_1 + y_2 + y_3 \\ x_2 = y_1 - y_2 \\ x_3 = y_1 + y_2 - 2y_3 \\ \end{array} \right. \Longleftrightarrow \left\{\begin{array}{l} y_1 = \frac{1}{3}x_1 + \frac{1}{2}x_2 + \frac{1}{6}x_3\\ y_2 = \frac{1}{3}x_1 - \frac{1}{2}x_2 + \frac{1}{6}x_3\\ y_3 = \frac{1}{3}x_1 - \frac{1}{3}x_3 \end{array} \right. \end{align} Then, \begin{align} y_1\ge y_2 \ge 0 \ge y_3 \Leftrightarrow \left\{\begin{array}{l} y_1 \ge y_2\\ y_2 \ge 0\\ 0 \ge y_3 \end{array} \right. \Leftrightarrow \left\{\begin{array}{l} \frac{1}{3}x_1 + \frac{1}{2}x_2 + \frac{1}{6}x_3 \ge \frac{1}{3}x_1 - \frac{1}{2}x_2 + \frac{1}{6}x_3\\ \frac{1}{3}x_1 - \frac{1}{2}x_2 + \frac{1}{6}x_3 \ge 0\\ 0 \ge \frac{1}{3}x_1 - \frac{1}{3}x_3 \end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x_2 \ge 0\\ 2x_1 + x_3 \ge 3x_2\\ x_3 \ge x_1 \\ \end{array} \right. \end{align}