Cho \(\Delta ABC\). Xét biểu thức \(S=\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\). Khẳng định nào sau đây đúng?
Giải chi tiết:
Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}\) ta có: \(\frac{1}{p-a}+\frac{1}{p-b}\ge \frac{4}{\left( p-a \right)+\left( p-b \right)}\Leftrightarrow \frac{1}{p-a}+\frac{1}{p-b}\ge \frac{4}{c}\,\,\,\,\,\,\,\,\,\,(1)\)
Tương tự:
\(\frac{1}{p-b}+\frac{1}{p-c}\ge \frac{4}{a}\,\,\,\,\,\,\,\,\,(2)\)
\(\frac{1}{p-c}+\frac{1}{p-a}\ge \frac{4}{b}\,\,\,\,\,\,\,\,\,(3)\)
\(\left( 1 \right)+\left( 2 \right)+\left( 3 \right)\Leftrightarrow 2\left( \frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c} \right)\ge 4\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\Leftrightarrow \frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge 2\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\)
Chọn D.