\(\frac{1}{1+2x}=1-\frac{1}{1+2y}+1-\frac{1}{1+2z}=\frac{2y}{1+1y}+\frac{2z}{1+2z}\ge4\sqrt{\frac{yz}{\left(1+2y\right)\left(1+2z\right)}}\)
Tương tự ta có: \(\frac{1}{1+2y}\ge4\sqrt{\frac{zx}{\left(1+2x\right)\left(1+2z\right)}}\); \(\frac{1}{1+2z}\ge4\sqrt{\frac{xy}{\left(1+2x\right)\left(1+2y\right)}}\)
Nhân vế với vế:
\(\frac{1}{\left(1+2x\right)\left(1+2y\right)\left(1+2z\right)}\ge\frac{64xyz}{\left(1+2x\right)\left(1+2y\right)\left(1+2z\right)}\)
\(\Rightarrow64xyz\le1\Rightarrow xyz\le\frac{1}{64}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{4}\)
