# Is there a more elegant way to prove this inequality?

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.

• Hint: Triangle inequality! $2 + x = (2x + 1) + (1 - x)$ Nov 3 '15 at 18:58
• How do I use this fact to prove this rigorously? Nov 3 '15 at 19:07
• Hint part 2: Let $a = 2x + 1$ and $b = 1 -x$. Apply the triangle inequality $|a + b| \leq |a| + |b|$ Nov 3 '15 at 19:07
• @SimonS - that should be an answer! Nov 3 '15 at 19:25

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