Question

Với a,b,c,d >0

Và \(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}+\dfrac{1}{1+d}\ge3\)

CMR: abcd \(\ge\) \(\dfrac{1}{81}\)

Answer 1

Lời giải:

Ta thấy:

\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=\frac{1}{a+1}+1-\frac{b}{b+1}+1-\frac{c}{c+1}+1-\frac{d}{d+1}\geq 3\)

\(\Rightarrow \frac{1}{a+1}\geq \frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\geq 3\sqrt[3]{\frac{bcd}{(b+1)(c+1)(d+1)}}\) (AM-GM)

Tương tự:

\(\frac{1}{b+1}\geq 3\sqrt[3]{\frac{acd}{(a+1)(c+1)(d+1)}}\)

\(\frac{1}{c+1}\geq 3\sqrt[3]{\frac{abd}{(a+1)(b+1)(d+1)}}\)

\(\frac{1}{d+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}\)

Nhân theo vế:

\(\Rightarrow \frac{1}{(a+1)(b+1)(c+1)(d+1)}\geq 81.\frac{abcd}{(a+1)(b+1)(c+1)(d+1)}\)

\(\Rightarrow abcd\leq \frac{1}{81}\)