Assume $\blue{w = u+iv}, \green{z = x+iy} \in \mathbb C$.
It is interesting to compare the two formulas.
In the Two Squares Equality, first we square each of the numbers,
then add pairs of the squares,
then multiply the resulting binary sums.
In the triangle inequality,
first we add matching components,
then square each binary sum,
then add the squares.
\[ \boxed{ \begin{array} {} \text{The Two Squares Equality} \\ \hline \blue w\green z = \blue{(u + iv)} \green{(x + iy)} = \blue u\green x - \blue v\green y +i(\blue v\green x + \blue u\green y) \\ \blue w \green{\bar z} = \blue{(u + iv)} \green{(x - iy)} = \blue u\green x + \blue v\green y + i(\blue v\green x - \blue u\green y) \\ \blue{\bar w} \green{\bar z} = \blue{(u - iv)}\green{(x - iy)} = \blue u\green x - \blue v\green y - i(\blue u\green y + \blue v\green x) = \overline{\blue w\green z} \\ \hline \blue{{|w|}^2} \green{{|z|}^2} = \blue w \blue{\bar w} \green{z \bar z} = \blue w \green {z \bar z} \blue{\bar w} = \blue w\green z \overline{\blue w\green z} = {|\blue w\green z|}^2 \\ \hline \boxed{ \blue{{|w|}^2} \green{{|z|}^2} = \blue{(u^2 + v^2)} \green{(x^2 + y^2)} } = \blue{u^2}\green{x^2} + \blue{u^2}\green{y^2} + \blue{v^2}\green{x^2} + \blue{v^2}\green{y^2} \\ {|\blue w\green z|}^2 = {(\blue u\green x - \blue v\green y)}^2 + {(\blue v\green x + \blue u\green y)}^2 \\ \blue{u^2}\green{x^2} - 2\blue u\green x\blue v\green y + \blue{v^2}\green{y^2} \\ \blue{v^2}\green{x^2} + 2\blue v\green x\blue u\green y + \blue{u^2}\green{y^2} \\ \end{array} } \]
Triangle inequality
\[ \boxed{ \begin{array} {} \text{The Triangle Inequality} \\ \hline |\blue W + \green Z| \leq \blue{|W|} + \green{|Z|} \\ {|\blue W+\green Z|}^2 \leq \blue{{|W|}^2} +2\blue{|W|}\green{|Z|} + \green{{|Z|}^2} \\ \boxed{{{|\blue W+\green Z|}^2 = (\blue u+\green x)}^2 + {(\blue v+\green y)}^2} \leq \blue{u^2 + v^2} + 2\blue{\sqrt{u^2 + v^2}}\green{\sqrt{x^2 + y^2}} + \green{x^2 + y^2} \\ 2(\blue u\green x+\blue v\green y) \leq 2\blue{\sqrt{u^2 + v^2}}\green{\sqrt{x^2 + y^2}} \\ {(\blue u\green x+\blue v\green y)}^2 \leq \blue{(u^2+v^2)} \green{(x^2+y^2)} \\ 2\blue u\green x\blue v\green y \leq \blue{u^2}\green{y^2} + \blue{v^2}\green{x^2} \\ 0 \leq \blue{u^2}\green{y^2} - 2\blue u\green x\blue v\green y + \blue{v^2}\green{x^2} = {(\blue u\green y - \blue v\green x)}^2 \\ \end{array} } \]
