Question

Cho a,b,c > 0. Chứng minh bất đẳng thức 1/a + 1/c +1/c >= 3((1/a+2b)+(1/b+2c)+(1/c+2a)

Answer 1

Áp dụng BĐT Svácxơ, ta có:

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

Tương tụ:

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

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

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

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

Dấu "=" xảy ra <=> \(a=b=c\)