
Inequality Proof
Hi I am working on my homework for Discrete Math and I don't know how to do this proof. It says, "Use proof by cases that x+y≤ x + y for all real numbers x and y." I know what proof by cases is but I don't know where to start.
Ok....I have started doing that and this is what I have so far:
Case 1: x + y>0
x+y>0 → x+y = x + y.
x + y cannot be > x + y.
∴ x + y ≤ x + y
Case 2: x + y = 0
x + y = 0 → x=x & y= x
∴ x + (x) ≤ x + y
Case 3: x + y < 0
x + y < 0 → x + y = (x + y).
I have gotten this far, but I don’t know where to go from here. I know it makes sense in my head, but I don’t know how to put it into words.

Here is another way.
$\displaystyle \begin{gathered}0 \leqslant x \Rightarrow \quad x = \left x \right \Rightarrow \quad x \leqslant \left x \right \hfill \\ x < 0 \Rightarrow \quad x < 0 <  x = \left x \right \hfill \\ \left( {\left {x + y} \right} \right)^2 =\left( {x + y} \right)^2 = x^2 + 2xy + y^2 \leqslant \left x \right^2 + 2\left x \right\left y \right + \left y \right^2 =\left( {\left x \right + \left y \right} \right)^2 \hfill \\\end{gathered} $
Now take the square root.

i understand the first two cases, but I don't understand the reason why you squared x + y in the last case.