Gọi AE cắt CD tại G. Dễ thấy \(\frac{AE}{AG}=\frac{BE}{BC}=\frac{3}{4},FG=DC\), do đó:
\(\frac{1}{2}AE.AF.\sin\widehat{EAF}=S_{AEF}=\frac{3}{4}S_{AFG}=\frac{3}{4}S_{ADC}=\frac{3}{8}AB.BC\)
Suy ra \(\sin\widehat{EAF}=\frac{3}{4}.\frac{AB.BC}{AE.AF}=\frac{3}{4}.\frac{xy}{\sqrt{x^2+\frac{9}{16}y^2}.\sqrt{y^2+\frac{1}{9}x^2}}\) \(\left(x=AB,y=BC\right)\)
\(\le\frac{3}{4}.\frac{xy}{xy+\frac{1}{4}xy}=\frac{3}{5}\) (BĐT Bunhiacopxki)
Vì \(0^0< \widehat{EAF}< 90^0\) nên \(max\widehat{EAF}=arc\sin\left(\frac{3}{5}\right)\approx36,87^0\)
Dấu "=" xảy ra khi và chỉ khi \(\frac{x}{y}=\frac{\frac{3}{4}y}{\frac{1}{3}x}\Leftrightarrow\frac{x}{y}=\frac{3}{2}\)hay \(\frac{AB}{BC}=\frac{3}{2}\Rightarrow\frac{AB}{AC}=\frac{3\sqrt{13}}{13}\)
