Question

Cho abc=1

CM: \(\dfrac{\text{1}}{\text{a}^2+2b^2+3}=\dfrac{\text{1}}{b^2+2c^2+3}=\dfrac{\text{1}}{c^2+2a^2+3}\) ≤ \(\dfrac{\text{1}}{\text{2}}\)

Answer 1

Sửa đề \("="\rightarrow"+"\)

Áp dụng BĐT cauchy, ta có:\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)

\(\Leftrightarrow\sum\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\right)\\ \Leftrightarrow\sum\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{ab^2c+abc+ab}+\dfrac{b}{abc+ab+b}\right)=\dfrac{1}{2}\cdot1=\dfrac{1}{2}\)

Dấu \("="\Leftrightarrow a=b=c=1\)