How to prove triangle inequality for euclidean distance on $\mathbb{R}^2$ without using Cauchy-Schwarz? 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?
 A: Let $x,y\in\mathbb{C}$. We want to prove that
$$
|x+y|\le |x|+|y|
$$
which is equivalent to
$$
|x+y|^2\le(|x|+|y|)^2
$$
Since $|x|^2=x\bar{x}$, the last inequality expands to
$$
|x|^2+x\bar{y}+\bar{x}y+|y|^2\le |x|^2+2|x||y|+|y|^2
$$
Note that $x\bar{y}+\bar{x}y$ is real and the inequality is certainly satisfied if $x\bar{y}+\bar{x}y\le0$. Suppose instead it is positive; then the inequality is the same as
$$
(x\bar{y}+\bar{x}y)^2\le 4|x|^2|y|^2
$$
Expanding and reordering yields
$$
(x\bar{y}-\bar{x}y)^2\le0
$$
which is true because $x\bar{y}-\bar{x}y$ is purely imaginary.
Now take $x=(a_1-z_1)+i(a_2-z_2)$ and $y=(z_1-b_1)+i(z_2-b_2)$ and you're done.
A: $x\mapsto x^2$ 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.
A: Suppose $u_1, u_2, c_1, c_2 \in\mathbb{R}.$
Then since $\ \left(\ u_1 (u_2 - c_2) - u_2 (u_1 - c_1)\ \right)^2 \geq 0,\ $ it follows that the equation:
$$ {u_1}^2 (u_2-c_2)^2  + {u_2}^2 (u_1-c_1)^2 \geq 2u_1 u_2 (u_1 - c_1) (u_2 - c_2)\qquad (1)$$
holds for all $u_1, u_2, c_1, c_2 \in\mathbb{R}.\ $ Now, $(1)\implies$
$$ \left( {u_1}^2 + {u_2}^2 \right) \left(\ (u_1-c_1)^2 + (u_2-c_2)^2\ \right) \geq \left(\ u_1 (u_1 - c_1) + u_2 ( u_2 - c_2)\ \right)^2 $$
$$ = \left(\ u_1 c_1 + u_2 c_2 - ({u_1}^2 + {u_2}^2 )\ \right)^2. $$
Square rooting both sides gives:
$$ \sqrt{ {u_1}^2 + {u_2}^2 } \sqrt{ (u_1-c_1)^2 + (u_2-c_2)^2 } \geq \sqrt{ \left(\ u_1 c_1 + u_2 c_2 - ({u_1}^2 + {u_2}^2 )\ \right)^2 } $$
$$ = \vert u_1 c_1 + u_2 c_2 - ({u_1}^2 + {u_2}^2 )\vert \geq u_1 c_1 + u_2 c_2 - ({u_1}^2 + {u_2}^2 ). $$
Multiplying through by $-2,\ $ adding $\ {c_1}^2 + {c_2}^2\ $ to both sides and rearranging gives:
$$ {c_1}^2 + {c_2}^2 \leq {u_1}^2 + {u_2}^2 + {u_1}^2 + {c_1}^2 - 2(c_1 u_1 + c_2 u_2) + {u_2}^2 + {c_2}^2$$ $$+ 2\sqrt{ {u_1}^2 + {u_2}^2 } \sqrt{ (u_1-c_1)^2 + (u_2-c_2)^2 }, $$
and now, noticing that the right-hand side is equal to $\ \left(\sqrt{ {u_1}^2 + {u_2}^2 } + \sqrt{ (u_1-c_1)^2 + (u_2-c_2)^2 }\right)^2,\ $ we have the result:
$$ {c_1}^2 + {c_2}^2 \leq \left(\sqrt{ {u_1}^2 + {u_2}^2 } + \sqrt{ (u_1-c_1)^2 + (u_2-c_2)^2 }\right)^2.\qquad (2)$$
Now, substituting $\ c_i = a_i - b_i,\ u_i = a_i - z_i\ $ for $\ i\in \{1,2\},\ $  and noticing that since we are working in $\mathbb{R},\ $ we have $\ \vert x\vert ^2 = x^2,\ \forall\ x\in\mathbb{R},\ $ we get the desired result:
$$\sqrt{|a_1-b_1|^2+|a_2-b_2|^2}\leq\sqrt{|a_1-z_1|^2+|a_2-z_2|^2}+\sqrt{|z_1-b_1|^2+|z_2-b_2|^2}.\qquad (3)$$
