\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
In the diagram, due to the triangular inequality, we must have \(AB \leq OA + OB\), and hence \(\abs *{z - w} \leq \abs *{z} + \abs *{w}\) as desired.
We have \begin {align*} \LHS & = \abs *{z - w}^2 \\ & = (z - w)(z - w)^* \\ & = (z - w)(z^* - w^*) \\ & = z z^* + w w^* - z w^* - z^* w \\ & = \abs *{z}^2 + \abs *{w}^2 - (E - 2\abs *{zw}) \\ & = \abs *{z}^2 + 2 \abs *{z} \abs *{w} + \abs *{w}^2 - E \\ & = (\abs *{z} + \abs *{w})^2 - E \\ & = \RHS , \end {align*}
exactly as desired.
Since \(\abs *{z - w}\), \(\abs *{z}\) and \(\abs *{w}\) are all real, so must be \(\abs *{z - w}^2\) and \((\abs *{z} + \abs *{w})^2\), and so \(E\) must be real.
Furthermore, we have \[ E = (\abs *{z} + \abs *{w})^2 - \abs *{z - w}^2, \] and by the inequality \(\abs *{z} + \abs *{w} \geq \abs *{z - w} \geq 0\), we can conclude \[ (\abs *{z} + \abs *{w})^2 \geq \abs *{z - w}^2, \] and hence \(E\) must be non-negative.
If we square both sides of the desired inequality (since both sides are non-negative this is reversible), we have \[ \frac {\abs *{z - w}^2}{\abs *{1 - zw^*}^2} \leq \frac {\left (\abs *{z} + \abs *{w}\right )^2}{\left (1 + \abs *{zw}\right )^2}, \] which is equivalent to showing \[ \frac {\left (\abs *{z} + \abs *{w}\right )^2 - E}{\left (1 + \abs *{zw}\right )^2 - E} \leq \frac {\left (\abs *{z} + \abs *{w}\right )^2}{\left (1 + \abs *{zw}\right )^2}. \]
We introduce a lemma. If \(a > c \geq 0\) and \(a > b\), then \[ \frac {b - c}{a - c} \leq \frac {b}{a}. \]
The proof of this is as follows. We cross-multiply the inequality to give (since \(a \geq a - c > 0\) this is reversible) \[ a(b - c) \leq b (a - c), \] which is equivalent to \[ ac \geq bc, \] and this must be true given \(c \geq 0\) and \(a > b\).
Now, since \(\abs *{z} > 1\), \(\abs *{w} > 1\), we have \[ (\abs *{z} - 1)(\abs *{w} - 1) = 1 + \abs *{zw} - \abs *{z} - \abs *{w} > 0, \] which means \[ 1 + \abs *{zw} > \abs *{z} + \abs *{w}, \] and since both are non-negative we have \[ (1 + \abs *{zw})^2 > (\abs *{z} + \abs *{w})^2. \]
Now, using this lemma, let \(a = (1 + \abs *{zw})^2, b = (\abs *{z} + \abs *{w})^2, c = E\). \(a > b\) is as shown in above, and \(c \geq 0\) is shown in part 1. \(a > c\) since \(a - c = \abs *{1 - zw^*}^2 \geq 0\), and the equal sign holds if and only if \(\abs *{zw^*} = \abs *{zw} = 1\), which must not hold if \(\abs *{z} > 1\) and \(\abs *{w} > 1\) since this gives \(\abs *{zw} = \abs *{z} \abs *{w} > 1\).
Therefore, we must have \[ \frac {\left (\abs *{z} + \abs *{w}\right )^2 - E}{\left (1 + \abs *{zw}\right )^2 - E} \leq \frac {\left (\abs *{z} + \abs *{w}\right )^2}{\left (1 + \abs *{zw}\right )^2}, \] which gives exactly what is desired.
This also holds for \(\abs *{z} < 1\) and \(\abs *{w} < 1\) since from this \((\abs *{z} - 1)(\abs *{w} - 1) > 0\) still holds, so \((1 + \abs *{zw})^2 > (\abs *{z} + \abs *{w})^2\) remains true, and \(\abs *{zw} = \abs *{z}\abs *{w} < 1\) so \(\abs *{zw} \neq 1\) remains true. The exact argument is still valid.