Question

  Cho a,b,c>0. Chứng minh \(\dfrac{ab}{a+3b+2c}\)+\(\dfrac{bc}{b+3c+2a}\)+\(\dfrac{ca}{c+3c+2b}\)≤\(\dfrac{a+b+c}{6}\)

   Mong mọi người giúp đỡ

 

Answer 1

Áp dụng BĐT

\(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\\ \Rightarrow\dfrac{9abc}{a+3a+2c}\\ =\dfrac{9}{\left(a+c\right)\left(b+c\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{4}{2}\) 

Tương tự với 2 BĐT còn lại rồi cộng vế theo vế

=> 9 vế trái

 \(\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\\ +\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{a+b+c}{2}\\ =\dfrac{3\left(a+b+c\right)}{2}\\ \Rightarrow......._{\left(đpcm\right)}\)