Proving inequality with change of variables I am working with this problem and I've come up with these inequalities. I have the next two equalities:
$y_1=\frac{1}{2}(x_1+x_2) $
$y_2=\frac{1}{2}(x_1-x_2)$
where the limits are:
$\qquad l_1\le x_1 \le u_1,$
$\qquad l_2\le x_2 \le u_2$
How can I prove that the limits of the variables $y_1,y_2$ are :
$\frac{1}{2}(l_1+l_2)\le y_1 \le \frac{1}{2}(u_1+u_2)$
$\frac{1}{2}(l_1-u_2)\le y_2 \le \frac{1}{2}(u_1-l_2)$
and  $x_1,x_2\in {\rm I\!R}$, 
$x_1,x_2 \ge 0$ 
I've done it graphically and I am certain that the feasible region of the two dimensional problem of having $x_1,x_2$ is a square or rectangle, depending on the upper and lower bounds. When I map these points into another Cartesian plane involving now $y_1,y_2$ this square or rectangle is rotated. I would like to know how this can be proven besides graphically.
Thank you so much in advance!
 A: How can we make $y_1=\frac12(x_1+x_2)$ as low as possible? We see that it is in a linear relationship with $x_1$ and $x_2$, such that if we lower $x_1$ or $x_2$ we lower $y_1$. It follows that the minimum is attained when $x_1,x_2$ are as low as possible – i.e. when we have $y_1=\frac12(l_1+l_2)$.
Similar reasoning applies to finding the upper bound of $y_1$ and the bounds of $y_2$. For $y_2$ also note that increasing $x_2$ will decrease (not increase) $y_2$ and vice versa.
A: Your claim can be proven using known properties of $\le$.
If $ l_1\le x_1 \le u_1$ and $ l_2\le x_2 \le u_2,$ 
then we can add these inequalities to get $  l_1+l_2\le x_1+x_2 \le u_1+u_2$ 
and divide by $2$ to get $\frac12( l_1+l_2)\le y_1 \le \frac12(u_1+u_2).$
Also $-u_2\le -x_2\le -l_{\,2},$ so, adding that to $l_1\le x_1\le u_1$, and dividing by $2,$
we have $\frac12(l_1-u_2)\le y_2\le \frac12 (u_1-l_2)$.
A: Just see the following steps:


*

*$ x_1\le u_1 \mbox{ and } x_2\le u_2 \Rightarrow x_1+x_2\le u_1+u_2$,

*but $x_1+x_2=2y_1$, then $2y_1\le u_1+u_2$.

*$ x_2\le u_2 \Rightarrow -u_2\le -x_2 $ and $l_1\le x_1 \mbox{ so } l_1-u_2\le x_1-x_2$ ,

*but $x_1-x_2=2y_1\Rightarrow l_1-u_2\le 2y_1$.


So you can conclude the rest as the same form
