Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{1-ab}=1+\dfrac{ab}{1-ab}\le1+\dfrac{ab}{1-\dfrac{a^2+b^2}{2}}=1+\dfrac{2ab}{a^2+b^2+2c^2}\)
\(=1+\dfrac{2ab}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}\le1+\dfrac{ab}{\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}}\)
\(\le1+\dfrac{1}{2}\left(\dfrac{a^2}{a^2+c^2}+\dfrac{b^2}{b^2+c^2}\right)\). Tương tự ta cũng có:
\(\dfrac{1}{1-bc}\le1+\dfrac{1}{2}\left(\dfrac{b^2}{a^2+b^2}+\dfrac{c^2}{a^2+c^2}\right);\dfrac{1}{1-ca}\le1+\dfrac{1}{2}\left(\dfrac{c^2}{b^2+c^2}+\dfrac{a^2}{a^2+b^2}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le3+\dfrac{1}{2}\left(\dfrac{a^2+b^2}{a^2+b^2}+\dfrac{b^2+c^2}{b^2+c^2}+\dfrac{c^2+a^2}{c^2+a^2}\right)=\dfrac{9}{2}\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)