Cho tam giác ABC nội tiếp (O;R). $AO\cap BC=\left \{ D \right \}$, $BO\cap AC=\left \{ E \right \}$, $CO\cap AB=\left \{ F \right \}$. CMR: $AD+BE+CF\geq \frac{9R}{2}$
Cho tam giác ABC nội tiếp (O;R). $AO\cap BC=\left \{ D \right \}$, $BO\cap AC=\left \{ E \right \}$, $CO\cap AB=\left \{ F \right \}$. CMR: $AD+BE+CF\geq \frac{9R}{2}$
Cho tam giác ABC nội tiếp (O;R). $AO\cap BC=\left \{ D \right \}$, $BO\cap AC=\left \{ E \right \}$, $CO\cap AB=\left \{ F \right \}$. CMR: $AD+BE+CF\geq \frac{9R}{2}$
Đặt $S_a=S_{OBC}, S_b=S_{OCA}, S_c=S_{OAB}, S=S_{ABC}$
Ta có:
$AD+BE+CF\geq \frac{9R}{2}$
$\Leftrightarrow 2(\frac{AD}{R}+\frac{BE}{R}+\frac{CF}{R})\geq 9$
$\Leftrightarrow 2(\frac{AD}{OA}+\frac{BE}{OB}+\frac{CF}{OC})\geq 9$
$\Leftrightarrow 2(\frac{S}{S_b+S_c}+\frac{S}{S_c+S_a}+\frac{S}{S_a+S_b})\geq 9$
$\Leftrightarrow 2S(\frac{1}{S_b+S_c}+\frac{1}{S_c+S_a}+\frac{1}{S_a+S_b})\geq 9$
$\Leftrightarrow [(S_b+S_c)+(S_c+S_a)+(S_a+S_b)]\left ( \frac{1}{S_b+S_c}+\frac{1}{S_c+S_a}+\frac{1}{S_a+S_b} \right )\geq 9$
Điều này đúng theo BĐT $Cauchy$
Dấu $"="$ xảy ra khi và chỉ khi $S_a=S_b=S_c$ hay $\Delta ABC$ đều
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