# General Proof for the triangle inequality

I am trying to prove: $P(n): |x_1| + \cdots + |x_n| \leq |x_1 + \cdots +x_n|$ for all natural numbers $n$. The $x_i$ are real numbers.

Base: Let $n =1$: we have $|x_1| \leq |x_1|$ which is clearly true

Step: Let $k$ exist in the integers such that $k \geq 1$ and assume $P(k)$ is true.

This is where I am lost. I do not see how to leverage the induction hypothesis.

Here is my latest approach:

Can you do the following in the induction step: Let $Y$ = |$x_1$ +...+$x_n$| and Let $Z$ = |$x_1$| +...+ |$x_n$| Then we have: |$Y$ + $x_n+1$| $\leq$ $Z$ + |$x_n+1$|. $Y$ $\leq$ $Z$ from the induction step, and then from the base case this is just another triangle inequality. End of proof.

• I think your inequality is wrong. $|1|+|-2|$ is not <= than $|-1+2|$.
– ivan
Sep 14, 2012 at 6:18
• The domain given is the natural numbers. Sep 14, 2012 at 6:18
• No. It says $n$ is natural. Otherwise this is an equality.
– ivan
Sep 14, 2012 at 6:21
• The domain of the index of $x$ is $\mathbb{N}$. It is not clear what the $x_k$ belong to? In any case, I would agree with @ivan's first comment. Sep 14, 2012 at 6:21
• I believe you have the inquality wrong and that is should be $\geq$ and not $\leq$ Sep 14, 2012 at 6:26

As @ivan indicates, the inequality is reversed - it should be

$$|x_1 + x_2 + \dots + x_n| \leq |x_1| + |x_2| + \dots + |x_n|$$

As the base case for induction, you need to show (or assert? can you take the "basic" triangle inequality for granted?)

$$|x_1 + x_2| \leq |x_1| + |x_2|.$$

Hint:

One way to do this is to show $(|a + b|)^2 \leq (|a| + |b|)^2$ by expanding the LHS and using $ab \leq |a||b|$.

Then, for induction, assume

$$|x_1 + x_2 + \dots + x_n| \leq |x_1| + |x_2| + \dots + |x_n|$$

and show

$$|x_1 + x_2 + \dots + x_n + x_{n+1}| \leq |x_1| + |x_2| + \dots + |x_n| + |x_{n+1}|$$

using the induction hypothesis and the base case.

• Can you do the following in the induction step: Let $Y$ = |$x_1$ +...+$x_n$| and Let $Z$ = |$x_1$| +...+ |$x_n$| Then we have: |$Y$ + $x_n+1$| $\leq$ $Z$ + |$x_n+1$|. $Y$ $\leq$ $Z$ from the induction step, and then from the base case this is just another triangle inequality. End of proof. Sep 15, 2012 at 19:00
• Right idea, though you should have $Y = x_1 + \dots + x_n$, (otherwise, when you write $|Y + x_{n+1}|$, you're referring to $||x_1 + \dots + x_n| + x_{n+1}|$. You'll have $|Y| \leq Z$ by the induction hypothesis. Sep 17, 2012 at 18:29

It is ok for $$|x_1 + x_2| \le |x_1| + |x_2|$$ (I)

$$|x_1 + x_2 +\cdots+ x_k| \le |x_1| + |x_2|+\cdots+ |x_k|$$ (Hypothesis)(II)

For $$n = k+1$$ :

a) 1st. Apply The triangle inequality for $$2$$ different "numbers" $$(x_1 + x_2 +\cdots+ x_k)$$ and $$x_{k+1}$$ because it is ok by (I)

b)2nd Apply the induction Hypothesis (II)

$$|x_1+x_2+\cdots+ x_k +x_{k+1}| \le |x_1 + x_2 +\cdots+x_k|+|x_{k+1}| \le |x_1|+|x_2|+\cdots+|x_k|+|x_{k+1}|$$