Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(\hept{\begin{cases}S_{OAB}=x\\S_{OBC}=y\\S_{OCA}=z\end{cases}}\)
Có: \(\frac{OA}{OD}=\frac{S_{AOB}}{S_{ODB}}=\frac{S_{AOC}}{S_{ODC}}=\frac{x+z}{y}\)
\(\Rightarrow\frac{R}{OD}=\frac{x+z}{y}\)
\(\Rightarrow OD=R.\frac{y}{x+z}\)
Tương tự, có: \(OD+OE+OF=R\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)
\(\ge\frac{3}{2}R.\)(BĐT Nesbitt)
A B C O D E F
Ta có: \(\frac{AD}{OD}=\frac{S\left(ABC\right)}{S\left(OBC\right)};\frac{BE}{OE}=\frac{S\left(BAC\right)}{S\left(OAC\right)};\frac{CF}{OF}=\frac{S\left(CBA\right)}{S\left(OBA\right)}\)
=> \(\frac{AD}{OD}+\frac{BE}{OE}+\frac{CF}{OF}=S\left(ABC\right)\left(\frac{1}{S\left(OBC\right)}+\frac{1}{S\left(OAC\right)}+\frac{1}{S\left(OAB\right)}\right)\)\(\ge S\left(ABC\right)\left(\frac{9}{S\left(OBC\right)+S\left(OAC\right)+S\left(OAB\right)}\right)=\frac{S\left(ABC\right).9}{S\left(ABC\right)}=9\)
=> \(\frac{AD}{OD}+\frac{BE}{OE}+\frac{CF}{OF}\ge9\)
=> \(\frac{AO+OD}{OD}+\frac{BO+OE}{OE}+\frac{CO+OF}{OF}\ge9\)
=> \(\frac{AO}{OD}+\frac{BO}{OE}+\frac{CO}{OF}\ge6\)
Dấu "=" xảy ra <=> \(S\left(OBC\right)=S\left(OAC\right)=S\left(OAB\right)\)