# Thread: Using triangle inequality to prove inequalities

1. ## Using triangle inequality to prove inequalities

I have been given the triangle inequality:

||x| - |y|| <= |x + y| <= |x| + |y|

where |x| is the absolute value of x.

Now I'm supposed to prove the following:

|3x + 2| + |3x - 2| >= 4 for all real x.

Now, I tried this:

|3x + 2| + |3x - 2| >= |3x + 2 + 3x - 2| = 6|x|.

This does not solve the problem. How do I do this?

2. $|3x + 2| = \begin{cases} 3x + 2 & x \geq - \frac{2}{3} \\ 2 - 3x & x < - \frac{2}{3}\end{cases}$

We can see that its minimum occurs at $x = -\frac{2}{3}$, i.e $|3x + 2| \geq 2$

We also find the same for $|3x - 2| \geq 2$

Conclusion follows

3. o_O: Sorry but that does not answer my question. I am supposed to use the triangle inequality to prove this.

4. Haha look at this ^^

$|3x-2|=|2-3x|$ by the property of absolute value, that is $|a|=|-a|$

Now consider $|3x+2|+|2-3x|$
this is in the form $|x|+|y|$, which is $\geq |x+y|$ by triangle inequality.

Hence $|3x+2|+|2-3x| \geq |3x+2+2-3x|=4$

5. Hahah, wow. How did I miss that? Thanks!