Question

Cho \(a+b+c=abc\)

Tìm GTNN của: \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)

Answer 1

\(abc=1\) chứ nhỉ?

Áp dụng bđt AM-GM:

\(\frac{1}{a^3\left(b+c\right)}+\frac{a\left(b+c\right)}{4}\ge\frac{1}{a}\)

\(\frac{1}{b^3\left(c+a\right)}+\frac{b\left(c+a\right)}{4}\ge\frac{1}{b}\)

\(\frac{1}{c^3\left(a+b\right)}+\frac{c\left(a+b\right)}{4}\ge\frac{1}{c}\)

\(\Rightarrow A+\frac{ab+bc+ac}{2}\ge ab+bc+ac\Rightarrow A\ge\frac{ab+bc+ac}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)

\("="\Leftrightarrow a=b=c=1\)