Question

Cho a,b,c \(\ge\) 0. Chứng minh rằng:

\(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)

Answer 1

\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge\dfrac{3}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Cộng vế với vế:

\(3\ge\dfrac{3+3\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

\(\Leftrightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)