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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 $$ \begin{align} &|x| \le |x+y| +|y| \\ &|x| - |y| \le |x + y| \end{align} $$

Martin Sleziak
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T. Webster
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2 Answers2

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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 $2=3-1>2-3=-1$

ant
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Yes your proof is correct. It might be possible to make it a bit shorter, but it works. However the first line of reasoning should be $|x+y| \leq |x| + |y|$, instead of $|x+y| - |y| \leq |x| + |y|$.

user2566092
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