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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 <=> \(S\left(OBC\right)=S\left(OAC\right)=S\left(OAB\right)\)
Đặt \(S_{BOC}=x^2,S_{AOC}=y^2,S_{AOB}=z^2\) \(\Rightarrow S_{ABC}=S_{BOC}+S_{AOC}+S_{AOB}=x^2+y^2+z^2\)
Ta có : \(\frac{AD}{OD}=\frac{S_{ABC}}{S_{BOC}}=\frac{AO+OD}{OD}=1+\frac{AO}{OD}=\frac{x^2+y^2+z^2}{x^2}=1+\frac{y^2+z^2}{x^2}\)
\(\Rightarrow\frac{AO}{OD}=\frac{y^2+z^2}{x^2}\Rightarrow\sqrt{\frac{AO}{OD}}=\sqrt{\frac{y^2+z^2}{x^2}}=\frac{\sqrt{y^2+z^2}}{x}\)
Tương tự ta có \(\sqrt{\frac{OB}{OE}}=\sqrt{\frac{x^2+z^2}{y^2}}=\frac{\sqrt{x^2+z^2}}{y};\sqrt{\frac{OC}{OF}}=\sqrt{\frac{x^2+y^2}{z^2}}=\frac{\sqrt{x^2+y^2}}{z}\)
\(\Rightarrow P=\frac{\sqrt{x^2+y^2}}{z}+\frac{\sqrt{y^2+z^2}}{x}+\frac{\sqrt{x^2+z^2}}{y}\ge\frac{x+y}{\sqrt{2}z}+\frac{y+z}{\sqrt{2}x}+\frac{x+z}{\sqrt{2}y}\)
\(=\frac{1}{\sqrt{2}}\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\right]\ge\frac{1}{\sqrt{2}}\left(2+2+2\right)=3\sqrt{2}\)
Dấu "=" xảy ra khi \(x=y=z\Rightarrow S_{BOC}=S_{AOC}=S_{AOB}=\frac{1}{3}S_{ABC}\)
\(\Rightarrow\frac{OD}{OA}=\frac{OE}{OB}=\frac{OF}{OC}=\frac{1}{3}\Rightarrow\)O là trọng tâm của tam giác ABC
Vậy \(MinP=3\sqrt{2}\) khi O là trọng tâm của tam giác ABC
a)
Ta có: \(\frac{OD}{AD}=\frac{S_{BOC}}{S_{ABC}};\frac{OE}{BE}=\frac{S_{AOC}}{S_{ABC}};\frac{OF}{CF}=\frac{S_{AOB}}{S_{ABC}}\)\(\Rightarrow\frac{OD}{AD}+\frac{OE}{BE}+\frac{OF}{CF}=\frac{S_{BOC}+S_{AOC}+S_{AOB}}{S_{ABC}}\)
\(\Rightarrow\frac{OD}{AD}+\frac{OE}{BE}+\frac{OF}{CF}=\frac{S_{ABC}}{S_{ABC}}=1\left(ĐPCM\right)\)
b) chịu
a. Đặt \(S_{AOB}=c^2;S_{BOC}=a^2;S_{COA}=b^2\Rightarrow S_{ABC}=a^2+b^2+c^2\)
Ta có \(\frac{AM}{OM}=\frac{S_{ABC}}{S_{BOC}}=\frac{a^2+b^2+c^2}{a^2}=1+\frac{b^2+c^2}{a^2}\)
Vậy thì \(\frac{OA}{OM}=\frac{AM}{OM}-1=\frac{b^2+c^2}{a^2}\Rightarrow\sqrt{\frac{OA}{OM}}=\sqrt{\frac{b^2+c^2}{a^2}}\ge\frac{1}{\sqrt{2}}\left(\frac{b}{a}+\frac{a}{b}\right)\)
Tương tự, ta có: \(\sqrt{\frac{OA}{OM}}+\sqrt{\frac{OB}{ON}}+\sqrt{\frac{OC}{OP}}\ge\frac{1}{\sqrt{2}}\left(\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}\right)\ge\frac{1}{\sqrt{2}}.6=3\sqrt{2}\)
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