\(% 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}}\)
Notice that by expanding this square, \begin {align*} \left (\sqrt {x_n} - \sqrt {y_n}\right )^2 & = x_n + y_n - 2 \sqrt {x_n y_n} \\ & = 2 a(x_n, y_n) - 2 g(x_n, y_n) \\ & = 2 (x_{n + 1} - y_{n + 1}). \end {align*}
Since this is a square, it must be non-negative, with the equal sign taking if and only if \(\sqrt {x_n} = \sqrt {y_n}\), which holds if and only if \(x_n = y_n\).
So \(x_{n + 1} \geq y_{n + 1}\), and \(x_{n + 1} = y_{n + 1}\) if and only if \(x_n = y_n\).
Since \(y_0 < x_0\), we have \(y_0 \neq x_0\), and hence \(y_1 \neq x_1\). By induction, this shows that \(y_n \neq _n\) for all \(n\), and hence for all \(n \geq 0\), \(y_n < x_n\).
Furthermore, \begin {align*} x_n - x_{n + 1} & = x_n - a(x_n, y_n) \\ & = x_n - \frac {x_n + y_n}{2} \\ & = \frac {x_n - y_n}{2} \\ & > 0, \end {align*}
since \(x_n > y_n\) and hence \(x_n > x_{n + 1}\).
Similarly, \begin {align*} y_{n + 1} - y_n & = g(x_n, y_n) - y_N \\ & = \sqrt {x_n y_n} - y_N \\ & = \sqrt {y_n} (\sqrt {x_n} - \sqrt {y_n}) \\ & > 0, \end {align*}
since \(x_n > y_n\) implies \(\sqrt {x_n} > \sqrt {y_n}\), and hence \(y_n < y_{n + 1}\).
Hence, for all \(n \in \NN \), \[ y_n < x_n < x_{n - 1} < x_{n - 2} < \cdots < x_0, \] and \(y_{n + 1} > y_n\).
Hence, \(\{y_N\}_{n = 0}^\infty \) is an increasing sequence, and is bounded above by \(x_0\).
So there exists \(M \in \RR \) such that \[ \lim _{n \to \infty } y_n = M. \]
As for the inequality, the left inequality sign is equivalent to \(y_{n + 1} < x_{n + 1}\) which was shown above.
To show the right inequality sign, this is equivalent to showing \begin {align*} \frac {1}{2} (\sqrt {x_n} - \sqrt {y_n})^2 & < \frac {1}{2} (x_n - y_n) \\ x_n + y_n - 2 \sqrt {x_n y_N} & < x_n - y_n \\ 2 y_n & < 2 \sqrt {x_n y_n} \\ \sqrt {y_n} & < \sqrt {x_n}, \end {align*}
which is true since \(y_n < x_n\).
Hence, \[ 0 < x_{n + 1} - y_{n + 1} < \frac {1}{2} (x_n - y_n) \] as desired.
Hence, we have \begin {align*} 0 & < x_n - y_n \\ & < \frac {1}{2} (x_{n - 1} - y_{n - 1}) \\ & < \frac {1}{4} (x_{n - 2} - y_{n - 2}) \\ & < \cdots \\ & < \frac {1}{2^n} (x_0 - y_0), \end {align*}
by induction.
\(x_0 - y_0 > 0\) is a positive real constant. Let \(n \to \infty \), and by the squeeze theorem, the strict inequalities become weak, and \[ 0 \leq \lim _{n \to \infty } (x_n - y_n) \leq \lim _{n \to \infty } \left (\frac {1}{2^n} (x_0 - y_0)\right ) = 0, \] and hence \[ \lim _{n \to \infty } (x_n - y_n) = 0. \]
Therefore, \begin {align*} \lim _{n \to \infty } x_n & = \lim _{n \to \infty } \left [(x_n - y_n) + y_n\right ] \\ & = \lim _{n \to \infty } (x_n - y_n) + \lim _{n \to \infty } y_n \\ & = 0 + M \\ & = M, \end {align*}
since both parts of the limit \(x_n - y_n\) and \(y_n\) exist, the limit of the sum is the sum of the limits of the individual parts.
Using this substitution, when \(x \to 0^{+}\), we have \(t \to -\infty \), and when \(x \to +\infty \), \(t \to +\infty \). Also, \[ \DiffFrac {t}{x} = \frac {1}{2} + \frac {1}{2} \cdot \frac {pq}{x^2} = \frac {1}{2} \left (1 + \frac {pq}{x^2}\right ). \]
Hence, the integral can be simplified as \begin {align*} & \phantom {=} \int _{-\infty }^{\infty } \frac {\Diff t}{\sqrt {\left (\frac {1}{4} (p + q)^2 + t^2\right ) \left (pq + t^2\right )}} \\ & = \int _{0}^{\infty } \frac {\frac {1}{2} \left (1 + \frac {pq}{x^2}\right ) \Diff x}{\sqrt {\left (\frac {1}{4} (p + q)^2 + \frac {1}{4} \left (x - \frac {pq}{x}\right )^2\right ) \left (pq + \frac {1}{4} \left (x - \frac {pq}{x}\right )^2\right )}} \\ & = \int _{0}^{\infty } \frac {\frac {1}{2} \left (1 + \frac {pq}{x^2}\right ) \Diff x}{\frac {1}{4} \sqrt {\left (p^2 + 2pq + q^2 + x^2 - 2pq + \frac {p^2 q^2}{x^2}\right ) \left (4pq + x^2 - 2pq + \frac {p^2 q^2}{x^2}\right )}} \\ & = 2 \int _{0}^{\infty } \frac {\left (1 + \frac {pq}{x^2}\right ) \Diff x}{\sqrt {\left (p^2 + q^2 + x^2 + \frac {p^2 q^2}{x^2}\right ) \left (x^2 + 2pq + \frac {p^2 q^2}{x^2}\right )}} \\ & = 2 \int _{0}^{\infty } \frac {(x^2 + pq) \Diff x}{\sqrt {\left (x^4 + (p^2 + q^2) x^2 + p^2 q^2\right )\left (x^4 + 2pq x^2 + p^2 q^2\right )}} \\ & = 2 \int _{0}^{\infty } \frac {(x^2 + pq) \Diff x}{\sqrt {(x^2 + p^2) (x^2 + q^2) (x^2 + pq)^2}} \\ & = 2 \int _{0}^{\infty } \frac {\Diff x}{\sqrt {(x^2 + p^2) (x^2 + q^2)}} \\ & = 2 I(p, q), \end {align*}
which means \[ \int _{-\infty }^{\infty } \frac {\Diff t}{\sqrt {\left (\frac {1}{4} (p + q)^2 + t^2\right ) \left (pq + t^2\right )}} = 2 I(p, q). \]
But also note that the left-hand side satisfies that \begin {align*} \LHS & = \int _{-\infty }^{\infty } \frac {\Diff t}{\sqrt {\left (\frac {1}{4} (p + q)^2 + t^2\right ) \left (pq + t^2\right )}} \\ & = 2 \int _{0}^{\infty } \frac {\Diff t}{\sqrt {\left [\left (\frac {1}{2} (p + q)\right )^2 + t^2\right ]\left [(\sqrt {pq})^2 + t^2\right ]}} \\ & = 2 \int _{0}^{\infty } \frac {\Diff t}{\sqrt {\left [a(p, q)^2 + t^2\right ]\left [g(p, q)^2 + t^2\right ]}} \\ & = 2 I(a(p, q), g(p, q)), \end {align*}
since the integrand is an even function, and so \[ I(p, q) = I(a(p, q), g(p, q)), \] as desired.
Since \(0 < q < p\), let \(y_0 = q, x_0 = p\), and hence \begin {align*} I(p, q) & = I(x_0, y_0) \\ & = I(a(x_0, y_0), g(x_0, y_0)) \\ & = I(x_1, y_1) \\ & = \cdots \\ & = I(x_n, y_n). \end {align*}
Let \(n \to \infty \), and we have \begin {align*} I(p, q) & = I(M, M) \\ & = \int _{0}^{\infty } \frac {\Diff x}{M^2 + x^2} \\ & = \frac {1}{M} \left [\arctan \left (\frac {x}{M}\right )\right ]_{0}^{\infty } \\ & = \frac {\pi }{2M}. \end {align*}