Question

[Xuân Lộc]

cho a,b,c>0 thoa man dieu kien a+b+c=1

c/m ab/c+1+bc/a+1+ac/b+1<=1/4

Answer 1

[Trần Nguyễn Bảo Quyên]

Áp dụng bất đẳng thức cho hai số dương

\(\dfrac{1}{\left(a+b\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Xét \(c+1=c+a+b+c\)

\(\dfrac{ab}{c+1}\le\dfrac{ab}{4\left[\dfrac{1}{a+c}+\dfrac{1}{b+c}\right]}\)

Tương tự:

\(\dfrac{bc}{a+1}\le\dfrac{bc}{4\left[\dfrac{1}{a+c}+\dfrac{1}{b+a}\right]}\)

\(\dfrac{ca}{b+1}\le\dfrac{ac}{4\left[\dfrac{1}{a+b}+\dfrac{1}{c+b}\right]}\)

Cộng lại :
\(\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\left[\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right]\)

Rút gọn mẫu số
\(\Rightarrow\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\)