BĐT cần chứng minh tương đương:
\(\frac{2}{2+a^2b}+\frac{2}{2+b^2c}+\frac{2}{2+c^2a}\ge2\)
\(\Leftrightarrow\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)
Ta có: \(VT=\sum\frac{a^2b}{1+1+a^2b}\le\frac{1}{3}\sum\frac{a^2b}{3\sqrt[3]{a^2b}}=\frac{1}{3}\sum\sqrt[3]{a^4b^2}=\frac{1}{3}\sum\sqrt[3]{a^2.ab.ab}\)
\(VT\le\frac{1}{9}\sum\left(a^2+ab+ab\right)=\frac{1}{9}\left(a+b+c\right)^2=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)