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 \(\dfrac{y^2+z^2}{x^2}\ge\dfrac{2yz}{x^2}\) ; \(\dfrac{x^2+y^2}{z^2}\ge\dfrac{2xy}{z^2}\)
\(\Rightarrow A=\dfrac{x^2+z^2}{y^2}.\dfrac{y^2+z^2}{x^2}.\dfrac{x^2+y^2}{z^2}\ge8\)
\(\sqrt{\dfrac{OA}{OP}}+\sqrt{\dfrac{OB}{OQ}}+\sqrt{\dfrac{OC}{OR}}\ge3\sqrt[3]{\sqrt{A}}=3\sqrt{2}\) - đpcm
