Say I have $x_1 > y_1 > 0$ and $x_{n+1} = \frac{x_n + y_n}{2}, y_{n+1} = \frac{2x_ny_n}{x_n + y_n}$ for $n \geq 1$. Show that $x_n > x_{n+1} > y_{n+1} > y_n > 0$.
So the obvious approach to me it seems is proceed by induction.
So with our base case, $n = 1$, $x_1 > \frac{x_1 + y_1}{2} > \frac{2x_1y_1}{x_1 + y_1} > y_1 > 0$.
Not sure if there's a good way to clean this up but not even sure how to show this holds true for the base case.
On
We don’t need induction indeed we have that
$$x_n > x_{n+1}\iff x_n =\frac{x_n + x_n}{2}>\frac{x_n + y_n}{2}$$
$$x_{n+1} >y_{n+1}\iff \frac{x_n + y_n}{2}> \frac{2x_ny_n}{x_n + y_n}\iff (x_n-y_n)^2>0$$
$$Y_{n+1} >y_{n}\iff \frac{2x_ny_n}{x_n + y_n}>y_n\iff \frac{x_n+x_n}{x_n + y_n}>1$$
On
Notice that for all $n$ we have $x_n,y_n>0$. Therefore if $$x_n>y_n>0$$ we have $$(x_n-y_n)^2>0\\(x_n+y_n)^2>4x_ny_n\\{x_n+y_n\over 2}>{2x_ny_n\over x_n+y_n}\\x_{n+1}>y_{n+1}$$since $x_1>y_1>0$ we have $x_n>y_n>0$ for all $n$. Also $$x_{n+1}={x_n+y_n\over 2}<{x_n+x_n\over 2}=x_n\\y_{n+1}={2x_ny_n\over x_n+y_n}$$since ${2au\over a+u}$ is an increasing function of $u$ for $a>0$ and $u\ge a$, the minimum is attained when $u=a$. Therefore $$y_{n+1}={2x_ny_n\over x_n+y_n}>{2y_ny_n\over y_n+y_n}=y_n$$from which we finally conclude that$$x_{n}>x_{n+1}>y_{n+1}>y_{n}$$
Your inductive hypothesis is $x_n > y_n$ then there are $3$ things to show firstly \begin{eqnarray*} x_n = \frac{x_n+x_n}{2} > \frac{x_n+y_n}{2} > x_{n+1}. \end{eqnarray*} Secondly AM-HM which follows from $(x_n-y_n)^2>0$ \begin{eqnarray*} \frac{x_n+y_n}{2} > \frac{2x_n y_n}{x_n+y_n}. \end{eqnarray*} Thirdly we have $x_n y_n > y_n^2$ now add $x_n y_n$ to both sides and divide by $x_n+y_n$ and we have \begin{eqnarray*} y_{n+1}= \frac{2x_n y_n}{x_n+y_n}> y_n. \end{eqnarray*}