Ta có:
\(\overrightarrow{AC}+\overrightarrow{BD}=\overrightarrow{AI}+\overrightarrow{IJ}+\overrightarrow{JC}+\overrightarrow{BI}+\overrightarrow{IJ}+\overrightarrow{JD}\)
\(=\left(\overrightarrow{AI}+\overrightarrow{BI}\right)+\left(\overrightarrow{JC}+\overrightarrow{JD}\right)+2\overrightarrow{IJ}=2\overrightarrow{IJ}\)
Lấy module 2 vế:
\(\left|\overrightarrow{AC}+\overrightarrow{BD}\right|=2\left|\overrightarrow{IJ}\right|\)
\(\Leftrightarrow\left|\overrightarrow{AC}+\overrightarrow{BD}\right|=2IJ\)
Mà theo BĐT vecto: \(AC+BD>\left|\overrightarrow{AC}+\overrightarrow{BD}\right|\)
\(\Rightarrow AC+BD>2IJ\)
