Question

giả sử a,b,c là các số thực dương thỏa mãn \(a^2+b^2+c^2=12\) chứng minh rằng

\(\dfrac{ab}{a+3b+2c}+\dfrac{bc}{b+3c+2a}+\dfrac{ca}{c+3a+2b}\)≤1

Answer 1

Áp dụng BĐT AM-Gm: ( dạng \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\))

\(VT=\sum\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\sum\dfrac{a}{2}+\sum\left[\dfrac{ab}{a+c}+\dfrac{bc}{a+c}\right]\right)\)

\(=\dfrac{1}{9}\left(\dfrac{a+b+c}{2}+a+b+c\right)=\dfrac{1}{6}\left(a+b+c\right)\)

\(\le\dfrac{1}{6}\sqrt{3\left(a^2+b^2+c^2\right)}=1\) (đpcm)

Dấu = xảy ra khi a=b=c=2