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Given the inequality: $$\left| 2+x \right| \le \left| 2x+1 \right| +\left| 1-x \right| $$

I want to prove that it will hold for every $x\in\mathbb{R}$

I know that I can go ahead and create $9$ different cases where each expression in the absolute values will be either greater than or less than $0$. However, I am wondering if there is a more elegant and efficient way to solve this. I think I se the triangle inequality in this inequality. What can I do?

Hints are much better appreciated than the actual solution.

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    $\begingroup$ Hint: Triangle inequality! $2 + x = (2x + 1) + (1 - x)$ $\endgroup$
    – Simon S
    Nov 3 '15 at 18:58
  • $\begingroup$ How do I use this fact to prove this rigorously? $\endgroup$
    – nikolita
    Nov 3 '15 at 19:07
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    $\begingroup$ Hint part 2: Let $a = 2x + 1$ and $b = 1 -x$. Apply the triangle inequality $|a + b| \leq |a| + |b|$ $\endgroup$
    – Simon S
    Nov 3 '15 at 19:07
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    $\begingroup$ @SimonS - that should be an answer! $\endgroup$ Nov 3 '15 at 19:25
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Have you proved the triangle inequality? It is the case that for all real numbers $a, b$,

$$|a + b| \leq |a| + |b|$$

Given that inequality (and the fact you do not need to reprove it) then setting $a = 2x + 1$ and $b = 1 -x$, we have

$$|x + 2| = |(2x + 1) + (1 - x)| \leq |2x + 1| + |1-x|$$


If you do not have a proof of the triangle inequality, search the site. I am sure it is out there.

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