Question

\(\left\{{}\begin{matrix}a,b,c\ge0\\\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}\ge2\end{matrix}\right.\)

CMR: \(abc\le\frac{1}{8}\)

Answer 1

Chắc bạn ghi nhầm đề, ko có số hạng \(\frac{1}{1+d}\)

\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge2\)

\(\Leftrightarrow\frac{1}{1+a}\ge1-\frac{1}{1+b}+1-\frac{1}{1+c}=\frac{b}{1+b}+\frac{c}{1+c}\ge2\sqrt{\frac{bc}{\left(1+b\right)\left(1+c\right)}}\)

Tương tự ta có:

\(\frac{1}{1+b}\ge2\sqrt{\frac{ca}{\left(1+c\right)\left(1+a\right)}}\) ; \(\frac{1}{1+c}\ge2\sqrt{\frac{ab}{\left(1+a\right)\left(1+b\right)}}\)

Nhân vế với vế:

\(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\frac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)

\(\Leftrightarrow abc\le\frac{1}{8}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)