Question

cho a,b,c >0 thỏa a+b+c=6 cm:
\(\dfrac{ab}{6+a-c}\)+\(\dfrac{bc}{6+b-a}\)+\(\dfrac{ca}{6+c-b}\)<=2

Answer 1

\(\dfrac{ab}{6+a-c}+\dfrac{bc}{6+b-a}+\dfrac{ca}{6+c-b}=\dfrac{ab}{2a+b}+\dfrac{bc}{2b+c}+\dfrac{ca}{2c+a}\)

Mà ta có:

\(\dfrac{2a+b}{ab}=\dfrac{2}{b}+\dfrac{1}{a}=\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{a}\ge\dfrac{9}{2b+a}\)

\(\Rightarrow\dfrac{ab}{2a+b}\le\dfrac{2b+a}{9}\)

Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{bc}{2b+c}\le\dfrac{2c+b}{9}\\\dfrac{ca}{2c+a}\le\dfrac{2a+c}{9}\end{matrix}\right.\)

Cộng 3 cái trên vế theo vế ta được

\(\dfrac{ab}{2a+b}+\dfrac{bc}{2b+c}+\dfrac{ca}{2c+a}\le\dfrac{3\left(a+b+c\right)}{9}=\dfrac{3.6}{9}=2\)