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.
 A: 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}
