Question

Cho 3 số dương a,b,c. CMR: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)

Answer 1

Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\) ta được

\(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{2b}\ge\dfrac{9}{2\left(a+2b\right)}\)

\(\dfrac{1}{2b}+\dfrac{1}{2c}+\dfrac{1}{2c}\ge\dfrac{9}{2\left(b+2c\right)}\)

\(\dfrac{1}{2c}+\dfrac{1}{2a}+\dfrac{1}{2a}\ge\dfrac{9}{2\left(c+2a\right)}\)

Cộng các BĐT theo vế

\(\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9}{2}\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)

Dấu " = " xảy ra khi a = b = c ( a,b,c > 0 )