Giải thích các bước giải:
a.Ta có:
$ABCD,CHIK$ là hình vuông $\to \widehat{HKC}=45^o=\widehat{ACD}$
$\to AC//HK$
Mà $ABCD$ là hình vuông $\to AC\perp BD\to HK\perp BD$
Lại có $BC\perp CD\to BC\perp DK\to H$ là trực tâm $\Delta DBK$
$\to DH\perp BK$
b.Ta có $H$ là trực tâm $\Delta BDK\to KN\perp BD, DM\perp BK, BC\perp DK$
$\to \widehat{DNK}=\widehat{DCB}=90^o$
Mà $\widehat{NDK}=\widehat{BDC}\to \Delta DNK\sim\Delta DCB(g.g)$
$\to \dfrac{DN}{DC}=\dfrac{DK}{DB}$
$\to DN.DB=DC.DK$
Tương tự $KM.KB=KC.KD$
$\to DN.BD+KM.BK=DC.DK+KC.KD=DK^2$
c.Gọi $S_{HBD}=S_1, S_{HDK}=S_2, S_{HBK}=S_3$
Ta có:
$\dfrac{HB}{HC}=\dfrac{S_{HBD}}{S_{HDC}}=\dfrac{S_{HBK}}{S_{HCK}}=\dfrac{S_{HBD}+S_{HBK}}{S_{HDC}+S_{HCK}}=\dfrac{S_1+S_3}{S_2}$
Tương tự ta có:
$\dfrac{DH}{HM}=\dfrac{S_1+S_2}{S_3}$
$\dfrac{KH}{HN}=\dfrac{S_2+S_3}{S_1}$
$\to \dfrac{HB}{HC}+\dfrac{HD}{HM}+\dfrac{HK}{HN}=\dfrac{S_1+S_3}{S_2}+\dfrac{S_1+S_2}{S_3}+\dfrac{S_2+S_3}{S_1}$
$\to \dfrac{HB}{HC}+\dfrac{HD}{HM}+\dfrac{HK}{HN}=\dfrac{S_1}{S_2}+\dfrac{S_3}{S_2}+\dfrac{S_1}{S_3}+\dfrac{S_2}{S_3}+\dfrac{S_2}{S_1}+\dfrac{S_3}{S_1}$
$\to \dfrac{HB}{HC}+\dfrac{HD}{HM}+\dfrac{HK}{HN}\ge 6\sqrt[6]{\dfrac{S_1}{S_2}\cdot\dfrac{S_3}{S_2}\cdot\dfrac{S_1}{S_3}\cdot\dfrac{S_2}{S_3}\cdot\dfrac{S_2}{S_1}\cdot\dfrac{S_3}{S_1}}$
$\to \dfrac{HB}{HC}+\dfrac{HD}{HM}+\dfrac{HK}{HN}\ge 6$
Dấu = xảy ra khi $S_1=S_2=S_3\to \Delta BDK$ đều (vô lý vì $\widehat{BDK}=45^o$ không đổi)
$\to$Dấu = không xảy ra
$\to \dfrac{HB}{HC}+\dfrac{HD}{HM}+\dfrac{HK}{HN}>6$
