Phương trình d theo đoạn chắn:
\(\dfrac{x}{m}+\dfrac{y}{n}=1\)
Thay tọa độ A vào ta được: \(\dfrac{1}{m}+\dfrac{2}{n}=1\Rightarrow\dfrac{1}{m}=1-\dfrac{2}{n}=\dfrac{n-2}{n}\)
\(\Rightarrow m=\dfrac{n}{n-2}\)
Mặt khác \(\left\{{}\begin{matrix}m>0\\n>0\end{matrix}\right.\Rightarrow\dfrac{n}{n-2}>0\Rightarrow n>2\)
\(\left\{{}\begin{matrix}OM=x_M=\dfrac{n}{n-2}\\ON=y_N=n\end{matrix}\right.\)
\(S_{OMN}=\dfrac{1}{2}OM.ON=\dfrac{n^2}{2\left(n-2\right)}=\dfrac{n+2}{2}+\dfrac{2}{n-2}=\dfrac{n-2}{2}+\dfrac{2}{n-2}+2\ge2\sqrt{\dfrac{2\left(n-2\right)}{2\left(n-2\right)}}+2=4\)
Dấu "=" xảy ra khi:
\(\dfrac{n-2}{2}=\dfrac{2}{n-2}\Leftrightarrow n=4\Rightarrow m=\dfrac{n}{n-2}=2\)
\(\Rightarrow m+n=6\)
