Problem about absolute value 
$$\begin{align*}
|x|=x &\text{if }x\geq 0\\
|x|=-x &\text{if }x\lt 0.
\end{align*}$$
  Show that $|xy|=|x||y|$.

I try to prove it as follows:
$|xy|=xy$ if $xy\geq 0$, 
but $xy\geq 0$ if and only if $x\geq 0$ and $y\geq 0$.
Since $x\geq 0$ and $y\geq 0$, then 
$x=|x|$ and $y=|y|$
hence from $|xy|=xy$
we have $|xy|=|x||y|$.
Am I correct?
How can I  show  the following two cases?
i. Show that $|x+y|\leq |x|+|y|$.
ii. If $y\gt 0$ and $-y\leq x \leq y$, then $|x|\lt y$
 A: *

*No, you are not correct. It is false that $xy\geq 0$ if and only if $x\geq 0$ and $y\geq 0$. For example, if $x=-1$ and $y=-1$, then $x\lt 0$, $y\lt 0$, and $xy = (-1)(-1) = 1\geq 0$. So your claim is incorrect, and thus your argument is incorrect. There are other possibilities, and you need to consider them too.

*Even if your argument were correct for that part, you would not be done yet. You only considered the case in which $xy\geq 0$. What if $xy\lt 0$? You also need to do that case.

*The last statement is incorrect. For example, if $x=y$, then $-y\leq x\leq y$ is true, but $|x|\lt y$ is false.

*To show (i), consider the different cases. What happens if $x$ and $y$ are both positive? If $x$ and $y$ are both negative? If $x$ is positive and $y$ negative and $x+y\geq 0$? If $x$ is positive, $y$ is negative, and $x+y\lt 0$? If $x$ is negative and $y$ positive? 

*To show (a corrected version of) (ii), consider the cases. What happens if $0\leq x\leq y$? What happens if $-y\lt x\lt 0$? 
