Question

cho \(\Delta\) ABC co cac canh a,b,c va chu vi 2p=a+b+c

chung minh \(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Answer 1

Ap dung bdt Cauchy-Schwarz dang Engel co:

\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{\left(1+1\right)^2}{p-a+p-b}=\dfrac{4}{2p-a-b}=\dfrac{4}{c}\)

Tuong tu: \(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\);

\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{b}\)

Cong theo ve cac bdt tren ta co:

\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

=> Đpcm