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enter image description here

plz, help me to write this equation , as arrangement form

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    Welcome to TeX SX! Could you post what you tried? – Bernard Mar 17 '18 at 00:59
  • And are you just wondering about the first inequality? Or the entire arrangement? – Teepeemm Mar 17 '18 at 01:37
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    I would recommend reading up on the AMS packages. It looks like the align environment would work well here (as well as your other question). – Teepeemm Mar 17 '18 at 01:39
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    It's possibly my eyes are deceiving me, but this looks like it was produced with eqnarray. You should avoid that. See https://tex.stackexchange.com/a/197/647 – TH. Mar 17 '18 at 03:53

1 Answers1

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\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}

enter image description here

g.kov
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