拆分长方程的写法

Question

请帮我把这个方程写成排列形式

Answer 1

\documentclass[10pt,a4paper]{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\begin{document}
First of all,
\begin{align}   
(1-\delta)^2(d(x_n,x_{n+1}))^2
&\le
\alpha^2 d(x_{n-1},x_n)d(x_n,x_{n+1})
+(\beta+\delta)^2(d(x_n,x_{n+1}))^2
\nonumber
\\
&\quad+
2\alpha(\beta+\delta)d(x_{n-1},x_n)
\sqrt{d(x_{n-1},x_n)d(x_n,x_{n+1})}
\\
\text{is equivalent to }\nonumber
\\
(1-\delta)^2(d(x_n,x_{n+1}))^2
&\le
\left(\alpha\sqrt{d(x_{n-1},x_n)d(x_n,x_{n+1})}
+(\beta+\delta)d(x_n,x_{n+1})\right)^2
.\label{1}
\end{align} 
We can also introduce a shortcut like,
let $d_k=d(x_k,x_{k+1})$. 

Then \eqref{1} becomes
\begin{align}   
((1-\delta) d_n)^2
&\le
\left(\alpha\sqrt{d_{n-1}d_n}
+(\beta+\delta)d_n\right)^2
.
\end{align}
Now if $d_n>d_{n-1}$, then
\begin{align}   
(1-\delta)^2 d_n^2
&\le
\left(\alpha\sqrt{d_nd_n}
+(\beta+\delta)d_n\right)^2
,\\
(1-\delta)^2 d_n^2
&\le
(\alpha
+\beta+\delta)^2 d_n^2
,\\
(1-\delta)^2
&\le
(\alpha
+\beta+\delta)^2
,\\
\dots\nonumber
\end{align}


\end{document}