Show inequality holds by induction

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.

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 $$\begin{eqnarray*} y_{n+1}= \frac{2x_n y_n}{x_n+y_n}> y_n. \end{eqnarray*}$$
$$x_n > x_{n+1}\iff x_n =\frac{x_n + x_n}{2}>\frac{x_n + y_n}{2}$$
$$x_{n+1} >y_{n+1}\iff \frac{x_n + y_n}{2}> \frac{2x_ny_n}{x_n + y_n}\iff (x_n-y_n)^2>0$$
$$Y_{n+1} >y_{n}\iff \frac{2x_ny_n}{x_n + y_n}>y_n\iff \frac{x_n+x_n}{x_n + y_n}>1$$
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$$x_{n}>x_{n+1}>y_{n+1}>y_{n}$$