Question

Cho hình bình hành ABCD có AC = 3 AD

C/M \(cot\widehat{BAD}\)\(\ge\dfrac{4}{3}\)

Answer 1

\(AC^2+BD^2=2\left(AB^2+AD^2\right)\)

\(\Leftrightarrow AB^2+AC^2=5BD^2\)

Áp dụng BĐT Cauchy:

\(cosBAD=\dfrac{AB^2+AD^2-BD^2}{2AB.AD}\ge\dfrac{4BD^2}{AB^2+AD^2}=\dfrac{4BD^2}{5BD^2}=\dfrac{4}{5}\)

\(\Rightarrow sinBAD=\sqrt{1-cos^2BAD}\le\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5}\)

\(\Rightarrow\dfrac{1}{sinBAD}\ge\dfrac{5}{3}\)

\(\Rightarrow cotBAD=\dfrac{cosBAD}{sinBAD}\ge\dfrac{4}{5}.\dfrac{5}{3}=\dfrac{4}{3}\)