Joseph Muscat's Functional Analysis asks the reader to prove $d(x,y)= \sqrt{|a_1-b_1|^2+|a_2-b_2|^2}$ on $\mathbb{R}^2$ satisfies the triangle inequality: $\sqrt{|a_1-b_1|^2+|a_2-b_2|^2}\leq\sqrt{|a_1-z_1|^2+|a_2-z_2|^2}+\sqrt{|z_1-b_1|^2+|z_2-b_2|^2}$ This is before the Cauchy-Schwarz inequality is explained.
I figured it would be the same as proving the triangle inequality for $|a_1-b_1|$ on $\mathbb{R}$, but I seriously don't know where to go from there to the inequality.
This was my attempt:
$d(x,y) = \sqrt{|a_1-b_1|^2+|a_2-b_2|^2} =\sqrt{|a_1-z_1+z_1-b_1|^2+|a_2-z_2+z_2-b_2|^2}$
$|a_1-z_1+z_1-b_1|\leq|a_1-z_1|+|z_1-b_1|$
So $\sqrt{|a_1-z_1+z_1-b_1|^2+|a_2-z_2+z_2-b_2|^2}\leq\sqrt{(|a_1-z_1|+|z_1-b_1|)^2+(|a_2-z_2|+|z_2-b_2|)^2}$
Expanding the right-hand side, we get:
$\sqrt{|a_1-z_1+z_1-b_1|^2+|a_2-z_2+z_2-b_2|^2}\leq\sqrt{(|a_1-z_1|^2+|z_1-b_1|^2+2|a_1-z_1||z_1-b_1|+|a_2-z_2|^2+|z_2-b_2|^2+2|a_2-z_2||z_2-b_2|}$
or
$\sqrt{|a_1-b_1|^2+|a_2-b_2|^2}\leq\sqrt{(|a_1-z_1|^2+|z_1-b_1|^2+2|a_1-z_1||z_1-b_1|+|a_2-z_2|^2+|z_2-b_2|^2+2|a_2-z_2||z_2-b_2|}$
I don't know where to go from here.
Is this approach reasonable?
Best Answer
$x\mapsto x^2$ is (strictly) increasing on $[0,\infty)$, i.e. $$a\leq b\quad\Leftrightarrow\quad a^2\leq b^2\ \qquad\forall a,b\geq 0\ .$$ So the triangle inequality is equivalent to $$d(x,y)^2\leq (d(x,z)+d(z,y))^2\ ,\qquad\forall x,y,z\in\mathbb R^2\ .$$ It is even sufficient to show that $${\rm(A)}:\quad d(x,y)^2\leq d(x,z)^2+d(z,y)^2\ ,\qquad\forall x,y,z\in\mathbb R^2\ ,$$ because $$d(x,z)^2+d(z,y)^2\leq d(x,z)^2+2d(x,z)d(z,y)+d(z,y)^2=(d(x,z)+d(z,y))^2\ ,\qquad\forall x,y,z\in\mathbb R^2\ .$$ Now you can use your estimations to show the inequality (A). By this way you avoid square roots, which is much easier than your original approach.
Note that the same techniques are used for the proof of the Cauchy-Schwarz inequality.