Question

1/a+3b + 1/b+3c + 1/c+3a => 1/a+2b+c + 1/b+2c+a + 1/c+2a+b 

cao nhân nào đó cứu tui giới làng nước ơi :)))

 

Answer 1

Áp dụng bđt buniacopxki dạng phân thức:

\(\dfrac{1}{a+3b}+\dfrac{1}{b+2c+a}\ge\dfrac{\left(1+1\right)^2}{2a+4b+2c}=\dfrac{2}{a+2b+c}\)

TT: \(\dfrac{1}{c+3a}+\dfrac{1}{a+2b+c}\ge\dfrac{2}{2a+b+c};\dfrac{1}{b+3c}+\dfrac{1}{c+2a+b}\ge\dfrac{2}{a+b+2c}\)

Cộng vế theo vế: 

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

<=>\(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\)

Dấu "=" <=> a=b=c