Có: \(abc=1\)
\(\Leftrightarrow a^2b^2c^2=1\)
Đặt \(A=\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(A=\frac{a^2b^2c^2}{a^3\left(b+c\right)}+\frac{a^2b^2c^2}{b^3\left(c+a\right)}+\frac{c^2a^2b^2}{c^3\left(a+b\right)}\)
\(A=\frac{b^2c^2}{a\left(b+c\right)}+\frac{a^2c^2}{b\left(c+a\right)}+\frac{a^2b^2}{c\left(a+b\right)}\)
Áp dụng BĐT Cauchy-schwarz dưới dạng engel ta có:
\(A\ge\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
Dấu " = " xảy ra <=> a=b=c=1
Áp dụng BĐT AM-GM ta có:
\(ab+bc+ac\ge3.\sqrt[3]{a^2b^2c^2}=3\sqrt[3]{1}=3\)
\(\Rightarrow A\ge\frac{3}{2}\)
Dấu " = " xảy ra <=> a=b=c=1
Tham khảo nhé~
