4
$\begingroup$

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

$\endgroup$
  • 4
    $\begingroup$ Probably easiest to start from $|x| = |x + y - y|$. $\endgroup$ – guy Sep 27 '13 at 19:59
  • 1
    $\begingroup$ @Trevor : using guy's hint you can get a much shorter proof: start with his hint, then $|x+y-y| \leq |x+y|+|-y| = |x+y|+|y|$, etc. $\endgroup$ – Stefan Smith Sep 27 '13 at 23:49
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $|x+y| - |y| \leq |x| + |y|$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.