Cách này hơi dài...gt$\Rightarrow\frac{4x}{5};\frac{4y}{5};\frac{4z}{5}\epsilon \left[ {0;1} \right]$
$\Rightarrow A=\frac{1}{\frac{4x}{5}+1}+\frac{2}{\frac{4y}{5}+1}+\frac{3}{\frac{4z}{5}+1}=5$
Use BĐT:$\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}\leq \frac{2}{1+ab}$
Mở rộng với 3 biến $a,b,c\epsilon \left[ {0;1} \right]$:
$\frac{1}{1+a^{3}}+\frac{1}{1+b^{3}}+\frac{1}{1+c^{3}}\leq \frac{3}{1+abc}$
$\Rightarrow \frac{1}{\frac{4x}{5}+1}+\frac{1}{\frac{4y}{5}+1}\leq \frac{2}{1+\frac{4}{5}\sqrt{xy}}$(1)
$\frac{1}{\frac{4y}{5}+1}+\frac{1}{\frac{4z}{5}+1}\leq \frac{2}{1+\frac{4}{5}\sqrt{yz}}$(2)
$\frac{2}{\frac{4z}{5}+1}+\frac{2}{\frac{4}{5}\sqrt{xy}+1}+\frac{2}{1+\frac{4}{5}\sqrt{yz}}\leq2.\frac{3}{1+\frac{4}{5}\sqrt[3]{z.\sqrt{xy}.\sqrt{yz}}}=\frac{6}{1+\frac{4}{5}\sqrt[6]{xy^{2}z^{3}}}$(3)
Từ(1)(2)(3)$\Rightarrow A\leq \frac{6}{1+\frac{4}{5}\sqrt[6]{xy^{2}z^{3}}}\Rightarrow P\leq (\frac{1}{4})^{6}$
Dấu''='' xra$\Leftrightarrow x=y=z=\frac{1}{4}$