\(\left(a+b\right)^2\le\left(a^2+bc\right)\left(1+\frac{b}{c}\right)=\frac{\left(a^2+bc\right)\left(b+c\right)}{c}\)
\(\Rightarrow\frac{1}{\left(a+b\right)^2}\ge\frac{c}{\left(a^2+bc\right)\left(b+c\right)}\)
Tương tự: \(\frac{1}{\left(a+c\right)^2}\ge\frac{b}{\left(a^2+bc\right)\left(b+c\right)}\)
Cộng vế với vế:
\(A\ge\frac{b+c}{\left(a^2+bc\right)\left(b+c\right)}+2\sqrt{a^2+bc}=\frac{1}{a^2+bc}+2\sqrt{a^2+bc}\)
\(A\ge\frac{1}{a^2+bc}+\sqrt{a^2+bc}+\sqrt{a^2+bc}\ge3\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{2}}\)
