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A B C M N P O
Ta có : \(\frac{OM}{AM}=\frac{S_{BOC}}{S_{ABC}}\) ; \(\frac{ON}{BN}=\frac{S_{AOC}}{S_{ABC}}\) ; \(\frac{OP}{CP}=\frac{S_{AOB}}{S_{ABC}}\)
\(\Rightarrow\frac{OM}{AM}+\frac{ON}{BN}+\frac{OP}{CP}=\frac{S_{ABC}}{S_{ABC}}=1\)
Áp dụng bđt Bunhiacopxki, ta có :
\(\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}=\left(\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}\right).\left(\frac{OM}{AM}+\frac{ON}{BN}+\frac{OP}{CP}\right)\ge\)
\(\ge\left(\sqrt{\frac{AM}{OM}.\frac{OM}{AM}}+\sqrt{\frac{BN}{ON}.\frac{ON}{BN}}+\sqrt{\frac{CP}{OP}.\frac{OP}{CP}}\right)^2=\left(1+1+1\right)^2=9\)
Vậy \(\frac{AM}{OM}+\frac{BN}{ON}+\frac{CP}{OP}\ge9\) (đpcm)
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: \(\sqrt{\frac{OA}{OM}}+\sqrt{\frac{OB}{ON}}+\sqrt{\frac{OC}{OP}}\ge\frac{1}{\sqrt{2}}\left(\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}\right)\ge\frac{1}{\sqrt{2}}.6=3\sqrt{2}\)
B A C O R Q P
Đặt \(S_{AOC}=x^2;S_{BOC}=y^2;S_{AOB}=z^2\) \(\left(x,y,z>0\right)\)
* Ta thấy tam giác AOB và BOP có chung đường cao kẻ từ B
\(\Rightarrow\dfrac{S_{AOB}}{S_{BOP}}=\dfrac{OA}{OP}\). Tương tự \(\dfrac{S_{AOC}}{S_{COP}}=\dfrac{OA}{OP}\)
\(\Rightarrow\dfrac{OA}{OP}=\dfrac{S_{AOB}}{S_{BOP}}=\dfrac{S_{AOC}}{S_{COP}}=\dfrac{S_{AOB}+S_{AOC}}{S_{BOP}+S_{COP}}=\dfrac{x^2+z^2}{y^2}\)
Tương tự \(\dfrac{OB}{OQ}=\dfrac{y^2+z^2}{x^2};\dfrac{OC}{OR}=\dfrac{x^2+y^2}{z^2}\)
* Áp dụng BĐT cau-chy ta có
\(\dfrac{x^2}{y^2}+\dfrac{z^2}{y^2}\ge2\sqrt{\dfrac{x^2z^2}{y^4}}=\dfrac{2xz}{y^2}\) .
Tương tự \(\dfrac{y^2+z^2}{x^2}\ge\dfrac{2yz}{x^2}\) ; \(\dfrac{x^2+y^2}{z^2}\ge\dfrac{2xy}{z^2}\)
\(\Rightarrow A=\dfrac{x^2+z^2}{y^2}.\dfrac{y^2+z^2}{x^2}.\dfrac{x^2+y^2}{z^2}\ge8\)* Áp dụng BĐT cauchy ta được
\(\sqrt{\dfrac{OA}{OP}}+\sqrt{\dfrac{OB}{OQ}}+\sqrt{\dfrac{OC}{OR}}\ge3\sqrt[3]{\sqrt{A}}=3\sqrt{2}\) - đpcm