TITLE: Help checking proof of reverse triangle inequality $|x| - |y| \le |x + y|$?
QUESTION [5 upvotes]: Let $x, y \in \mathbb{R}$. Prove $|x| - |y| \le |x + y|$. 
By the the triangle inequality $|x| + |y| \ge |x + y|$, hence 
$$
\begin{align}
&|y| \ge |x+y| - |x| \\
&|x+y| \ge |x+y| - |y| \\
\end{align}
$$
Subtracting the first inequality from the second, we have
$$
\begin{align}
&|x+y|-|y| \ge |x+y| - |y| - (|x+y| - |x|) = |x| - |y| \\
&|x+y| \ge |x+y| - |y| \ge |x| - |y| \\
&|x+y| \ge |x|- |y| \\
\end{align}
$$ 
The mistake in this proof is that subtracting the inequalities isn't generally valid. For example, $101 \ge 100$ and $100 \ge 1$ but $101 - 100 = 1 \not\ge 99$.
Taking someone's hint, a shorter proof is 
$$|x| = |x + y - y| = |x + y + (-y)| \le |x + y| + |-y| = |x + y| + |y| $$
by the triangle inequality. Then
$$
\begin{align}
&|x| \le |x+y| +|y| \\
&|x| - |y| \le |x + y|
\end{align}
$$

REPLY [3 votes]: Sorry, I can't do comments yet.
Your first line after subtracting inequalities is incorrect.
$$|x+y|−|y|≥|x|−|y|$$
if $x=1/2$, $y=-1/2$ we get $-1/2 \ge 0$
We can't subtract inequalities like that, the inequality we are subtracting will be reversed.
e.g. $3>2$ and $3>1$ so $0=3-3 \not> 2-1=1$ 
but $2=3-1>2-3=-1$