\(\Sigma\frac{1}{6+a}\ge\frac{1}{2}\)
\(\Rightarrow\frac{1}{6+a}\ge\left(\frac{1}{6}-\frac{1}{6+b}\right)+\left(\frac{1}{6}-\frac{1}{6+c}\right)+\left(\frac{1}{6}-\frac{1}{6+d}\right)\)
\(=\frac{b}{6\left(6+b\right)}+\frac{c}{6\left(6+c\right)}+\frac{d}{6\left(6+d\right)}\ge3\sqrt[3]{\frac{bcd}{6\left(6+b\right).6\left(6+c\right).6\left(6+d\right)}}\)
\(=\frac{1}{2}\sqrt[3]{\frac{bcd}{\left(6+b\right)\left(6+c\right)\left(6+d\right)}}\)
Tương tự ta có :
\(\frac{1}{6+b}\ge\frac{1}{2}\sqrt[3]{\frac{acd}{\left(6+a\right)\left(6+c\right)\left(6+d\right)}}\)
\(\frac{1}{6+c}\ge\frac{1}{2}\sqrt[3]{\frac{abd}{\left(6+a\right)\left(6+b\right)\left(6+d\right)}}\)
\(\frac{1}{6+d}\ge\frac{1}{2}\sqrt[3]{\frac{abc}{\left(6+a\right)\left(6+b\right)\left(6+c\right)}}\)
Nhận các vế với nhau ta được :
\(\frac{1}{\left(6+a\right)\left(6+b\right)\left(6+c\right)\left(6+d\right)}\ge\frac{1}{16}.\sqrt[3]{\left(\frac{abcd}{\left(6+a\right)\left(6+b\right)\left(6+c\right)\left(6+d\right)}\right)^3}\)
\(\Rightarrow\frac{abcd}{16}\le1\)
\(\Rightarrow abcd\le16\)
Dấu " = " xảy ra khi \(a=b=c=d=2\)
Chúc bạn học tốt !!
\(\frac{1}{6+a}\ge\frac{1}{6}-\frac{1}{6+b}+\frac{1}{6}-\frac{1}{6+c}+\frac{1}{6}-\frac{1}{6+c}\)
\(\frac{1}{6+a}\ge\frac{b}{6\left(6+b\right)}+\frac{c}{6\left(6+c\right)}+\frac{d}{6\left(6+d\right)}\ge\frac{1}{2}\sqrt[3]{\frac{bcd}{\left(6+a\right)\left(6+b\right)\left(6+c\right)}}\)
Tương tự: \(\frac{1}{6+b}\ge\frac{1}{2}\sqrt[3]{\frac{acd}{\left(6+a\right)\left(6+c\right)\left(6+d\right)}}\) ; \(\frac{1}{6+c}\ge\frac{1}{2}\sqrt[3]{\frac{abd}{\left(6+a\right)\left(6+b\right)\left(6+d\right)}}\)
\(\frac{1}{6+d}\ge\frac{1}{2}\sqrt[3]{\frac{abc}{\left(6+a\right)\left(6+b\right)\left(6+c\right)}}\)
Nhân vế với vế và rút gọn:
\(1\ge\frac{abcd}{16}\Rightarrow abcd\le16\)
Dấu "=" xảy ra khi \(a=b=c=d=2\)
