Question

Answer 1

Đặt vế trái là P, ta có:

\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+b}{8}+\dfrac{1+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)\left(1+c\right)}{64\left(1+b\right)\left(1+c\right)}}=\dfrac{3a}{4}\)

Tương tự:

\(\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{1+c}{8}+\dfrac{1+a}{8}\ge\dfrac{3b}{4}\)

\(\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{1+a}{8}+\dfrac{1+b}{8}\ge\dfrac{3c}{4}\)

Cộng vế:

\(P+\dfrac{6+2a+2b+2c}{8}\ge\dfrac{3a+3b+3c}{4}\)

\(\Rightarrow P\ge\dfrac{a+b+c}{2}-\dfrac{3}{4}\ge\dfrac{3\sqrt[3]{abc}}{2}-\dfrac{3}{4}=\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(a=b=c=1\)