\(\dfrac{1}{1+2x}+\dfrac{1}{1+2y}+\dfrac{1}{1+2z}=2\Leftrightarrow\dfrac{1}{1+2x}=1-\dfrac{1}{1+2y}+1-\dfrac{1}{1+2z}\Leftrightarrow\dfrac{1}{1+2x}=\dfrac{2y}{1+2y}+\dfrac{2z}{1+2z}\)
Áp dụng bđt cosi ta có
\(\dfrac{2y}{1+2y}+\dfrac{2z}{1+2z}\ge2\sqrt{\dfrac{2y.2z}{\left(1+2y\right)\left(1+2z\right)}}=\dfrac{4\sqrt{yz}}{\sqrt{\left(1+2y\right)\left(1+2z\right)}}\Leftrightarrow\dfrac{1}{1+2x}\ge\dfrac{4\sqrt{yz}}{\sqrt{\left(1+2y\right)\left(1+2z\right)}}\)(1)
Chứng minh tương tự: \(\dfrac{1}{1+2y}\ge\dfrac{4\sqrt{xz}}{\sqrt{\left(1+2x\right)\left(1+2z\right)}}\)(2)
\(\dfrac{1}{1+2z}\ge\dfrac{4\sqrt{xy}}{\sqrt{\left(1+2x\right)\left(1+2y\right)}}\)(3)
Lấy (1).(2).(3) theo vế ta được
\(\dfrac{1}{1+2x}.\dfrac{1}{1+2y}.\dfrac{1}{1+2z}\ge\dfrac{4\sqrt{yz}}{\sqrt{\left(1+2y\right)\left(1+2z\right)}}.\dfrac{4\sqrt{xz}}{\sqrt{\left(1+2x\right)\left(1+2z\right)}}.\dfrac{4\sqrt{xy}}{\sqrt{\left(1+2x\right)\left(1+2y\right)}}=\dfrac{64xyz}{\left(1+2x\right)\left(1+2y\right)\left(1+2z\right)}\Leftrightarrow64xyz\le1\Leftrightarrow xyz\le\dfrac{1}{64}\Leftrightarrow P\le\dfrac{1}{64}\)
Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{1+2y}=\dfrac{1}{1+2z}\\\dfrac{1}{1+2z}=\dfrac{1}{1+2x}\\\dfrac{1}{1+2x}=\dfrac{1}{1+2y}\\xyz=\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow x=y=z=\dfrac{1}{4}\)
Vậy GTLN của P=\(\dfrac{1}{64}\)
