Euclidean Metric on Real Vector Space is Metric

Theorem
The Euclidean metric is a metric.

Proof
It is easy to see that conditions M0, M1 and M2 of the conditions for being a metric are satisfied. So all we need to do is check M3.

Let:
 * 1) $$z = \left({z_1, z_2, \ldots, z_n}\right)$$;
 * 2) all summations be over $$i = 1, 2, \ldots, n$$;
 * 3) $$x_i - y_i = r_i$$;
 * 4) $$y_i - z_i = s_i$$.

Then:

$$\left({\sum \left({x_i - y_i}\right)^2}\right)^{1 / 2} + \left({\sum \left({y_i - z_i}\right)^2}\right)^{1 / 2} = \left({\sum r_i^2}\right)^{1 / 2} + \left({\sum s_i^2}\right)^{1 / 2}$$

So we have to prove:

$$\left({\sum r_i^2}\right)^{1 / 2} + \left({\sum s_i^2}\right)^{1 / 2} \ge \left({\sum \left({r_i + s_i}\right)^2}\right)^{1 / 2}$$

Now:

$$ $$

... which we get by squaring both sides. The inequality continues to hold for both equations because both sides are non-negative.

Next, we have:

$$ $$

... which leads us to:

$$ $$ $$

This is just Cauchy's Inequality. Job done.