Exercise 2.4.6

Question

[Arithmetic-Geometric Mean]
  \begin{enumerate}[label=(\alph*)] 
  \item Explain why $\sqrt{x y} \leq$ $(x+y) / 2$ for any two positive real numbers $x$ and $y$. (The geometric mean is always less than the arithmetic mean.)
  \item Now let $0 \leq x_{1} \leq y_{1}$ and define
    $$
    x_{n+1}=\sqrt{x_{n} y_{n}} \quad 	ext { and } \quad y_{n+1}=\frac{x_{n}+y_{n}}{2}
    $$
    Show $\lim x_{n}$ and $\lim y_{n}$ both exist and are equal.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We have
    $$
    \sqrt{xy} \le (x+y)/2 \iff 4xy \le x^2 + 2xy + y^2 \iff 0 \le (x - y)^2
    $$
  \item The only fixed point is $x_n = y_n$ so we only need to show both sequences converge.

The inequality $x_1 \le y_1$ is always true since
    $$
    \sqrt{x_ny_n} \le \frac{x_n + y_n}{2} \implies x_{n+1} \le y_{n+1}
    $$
    Also $x_n \le y_n$ implies $(x_n+y_n)/2 = y_{n+1} \le y_n$, similarly $\sqrt{x_ny_n} = x_{n+1} \ge x_n$ meaning both sequences converge by the monotone convergence theorem.
   \end{enumerate}

Exercise 2.4.6

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Exercise 2.4.6

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