I need to prove that for all $n \in \mathbb{N}$ we have the inequality $$ \sqrt{\sum_{k=1}^n (x_k - y_k)^2} \leq \sqrt{\sum_{k=1}^n (x_k - z_k)^2} + \sqrt{\sum_{k=1}^n (z_k - y_k)^2}. $$

The hint says this follows from Cauchy Schwarz, but I don't see how. Cauch-Schwarz says we have $$| \langle x, y\rangle | \leq || x || \cdot || y ||. $$ How can I use that here?


1 Answer 1


That is the triangle inequality in $\mathbb R^n$! In general you can prove the triangle inequality using Cauchy-Schwarz in any space where the norm arises from the inner product (that is, $||x||^2 = \langle x,x\rangle$; clearly $\mathbb R^n$ is such a space). See for example the wikipedia entry on triangle inequality.

Now let $||\cdot||$ be the usual norm in $\mathbb R^n$ and $(\cdot, \cdot)$ be the usual inner product. Then you can show it as follows:

$$||x+y||^2 = (x+y,x+y) \le ||x||^2 + ||y||^2 + 2|(x,y)|\le ||x||^2 + ||y||^2 + 2||x||\cdot ||y|| = (||x|| + ||y||)^2$$ where we used the Cauchy Scwarz inequality on the penultimate step.

So we showed $||x+y|| \le ||x|| + ||y||$; hence you can write

$$||x-y|| = ||x-z + z-y|| \le ||x-z|| + ||z-y||$$

which is what you wanted to prove.

Note that complete metric spaces where the norm is generated by the inner product are called Hilbert spaces. They are especially important and used everywhere


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