Áp dụng BĐT \(Cauchuys-Schwarz\), ta có:
\(\dfrac{1}{a^2\left(1+bc\right)}=\dfrac{1}{a^2+a^2bc}=\dfrac{1}{a^2+a\left(a+b+c\right)}=\dfrac{1}{a^2+ab+a^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{a^2+ab}+\dfrac{1}{a^2+ac}\right)\)
\(\dfrac{1}{b^2\left(1+ca\right)}\le\dfrac{1}{4}\left(\dfrac{1}{b^2+ab}+\dfrac{1}{b^2+bc}\right)\)
\(\dfrac{1}{c^2\left(1+ab\right)}\le\dfrac{1}{4}\left(\dfrac{1}{c^2+ac}+\dfrac{1}{c^2+bc}\right)\)
Cộng tất cả vế, ta được:
\(\dfrac{1}{a^2\left(1+bc\right)}+\dfrac{1}{b^2\left(1+ca\right)}+\dfrac{1}{c^2\left(1+ab\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{a\left(a+c\right)}+\dfrac{1}{b\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(a+c\right)}+\dfrac{1}{c\left(b+c\right)}\right)\)
\(\le\dfrac{1}{4}\left(\dfrac{b+a}{ab\left(a+b\right)}+\dfrac{c+a}{ac\left(a+c\right)}+\dfrac{c+b}{bc\left(b+c\right)}\right)\)
\(\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)
\(\le\dfrac{1}{4}\left(\dfrac{a+b+c}{abc}\right)\)
\(\le\dfrac{1}{4}\left(\dfrac{abc}{abc}\right)=\dfrac{1}{4}\)
Vậy \(\dfrac{1}{a^2\left(1+bc\right)}+\dfrac{1}{b^2\left(1+ca\right)}+\dfrac{1}{c^2\left(1+ab\right)}\le\dfrac{1}{4}\)
