I need help verifying my attempted proof of the lower triangle inequality which states that
\begin{align}
\text{Let $x,y\in \mathbb{R}$, $||x|-|y|| \leq |x-y|$}
\end{align}
I used is the theorem which states that
\begin{align}
\text{Let $a \geq 0$.$|x| \leq a$ iff $-a \leq x \leq a$ where $x,a \in \mathbb{R}$}
\end{align}
The following is the proof
Let $x,y$ be real numbers, $|x| – |y| \leq |x-y|$
Proof. In case $x = y = 0$,
\begin{align}
0 = 0
\end{align}
establishing the result for this case.
In case that $x = 0$ and $y$ is a non-zero real number,
\begin{align}
-(-y) < -y \text{ for $y<0$};\\
-(-y) = -(-y) \text{ for $y>0$}
\end{align}
establishing the result for the case.
In case that $x$ is a non-zero and $y=0$
\begin{align}
-x = -x \text{ for $x<0$};\\
x = x \text{ for $x>0$}
\end{align}
establishing the result for this case.
In case that $x,y$ are non-zero and $x>y$,
\begin{align}
x-y = x-y &\text{ for $x,y>0$};\\
x-(-y) < x-y &\text{ for $x>0$, $y<0$, and $|x|<|y|$ (the $LHS$ is negative)};\\
x-(-y) < x-y &\text{ for $x>0$, $y<0$, and $|x|>|y|$ (the $LHS$ is positive)};\\
(-x)-(-y) < |x-y| &\text{ for $x,y<0$ (the $LHS$ is negative)}
\end{align}
establishing the results for this case.
In case that $x,y$ are non-zero and $x<y$,
\begin{align}
x-y < |x-y| &\text{ for $x,y>0$ (the $LHS$ is negative)};\\
(-x)-y < |x-y| &\text{ for $x<0$, $y>0$, and $|x|<|y|$ (the $LHS$ is negative)};\\
(-x)-y < |x-y| &\text{ for $x<0$, $y>0$, and $|x|>|y|$ (the $LHS$ is positive)};\\
(-x)-(-y) = |x-y| &\text{ for $x,y<0$}
\end{align}
establishing the result for this case.
Since the result for each possible cases has been established, $|x|-|y| \leq |x-y|$.
Let $x,y$ be real numbers, $|x| – |y| \geq -|x-y|$
Proof. In case $x = y = 0$,
\begin{align}
0 = 0
\end{align}
establishing the result for this case.
In case that $x = 0$ and $y$ is a non-zero real number,
\begin{align}
-|y| \geq -|-y|\\
\text{Because $|-y| = |-1||y| = |y|$}\\
-|y| = -|y|
\end{align}
establishing the result for the case.
In case that $x$ is a non-zero and $y=0$
\begin{align}
|x| \geq -|x|
\end{align}
establishing the result for this case.
In case that $x,y$ are non-zero and $x>y$,
\begin{align}
x-y > -|x-y| &\text{ for $x,y>0$};\\
(x)-(-y) > -|x-y| &\text{ for $x>0$, $y<0$, and $|x|<|y|$ (the $LHS$ is negative)};\\
(x)-(-y) > -|x-y| &\text{ for $x>0$, $y<0$, and $|x|>|y|$ (the $LHS$ is positive)};\\
(-x)-(-y) = -|x-y| &\text{ for $x,y<0$}
\end{align}
establishing the results for this case.
In case that $x,y$ are non-zero and $x<y$,
\begin{align}
x-y = -|x-y| &\text{ for $x,y>0$};\\
(-x)-y > -|x-y| &\text{ for $x<0$, $y>0$, and $|x|<|y|$ (the $LHS$ is negative)};\\
(-x)-y > -|x-y| &\text{ for $x<0$, $y>0$, and $|x|>|y|$ (the $LHS$ is positive)};\\
(-x)-(-y) > -|x-y| &\text{ for $x,y<0$}
\end{align}
establishing the result for this case.
Since the result for each possible cases has been established, $|x|-|y| \geq -|x-y|$.
Because $-|x-y| \leq |x|-|y| \leq |x-y|$, by the theorem in the beginning of this post, $||x|-|y|| \leq |x-y|$
I think my attempted proof is a conjecture rather than a proof.
Reference
Daepp, U. and Gorkin, P., 2011. Reading, Writing, and Proving. 2nd ed. pp. 55.
Best Answer
Here it is another way to approach it for the sake of curiosity.
On one hand, we have that
\begin{align*} |x| = |(x - y) + y| \leq |x - y| + |y| \Rightarrow |x - y| \geq |x| - |y| \end{align*}
On the other hand, we have that \begin{align*} |y| = |(y - x) + x| \leq |y - x| + |x| \Rightarrow |x -y | \geq |y| - |x| \end{align*}
Gathering both results, it results that \begin{align*} |x - y| \geq \max\{|x| - |y|,-(|x| - |y|)\} = ||x| - |y|| \end{align*}
and we are done.