1) áp dụng BĐT $a^3+b^3+c^3 \geq \frac{(a+b+c)^3}{9}$
=> $P^3 \leq 9(x+3y+y+3z+z+3x)=27$
=>max P=3 khi x+3y=y+3z=z+3x
<=>x=y=z=1/4
2)
mặt khác $6=\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{z} \geq 6\sqrt[6]{\frac{1}{xy^2z^3}}$
<=> $xy^2z^3 \geq 1$
M$=x+y^2+z^3 \geq 3\sqrt[3]{xy^2z^3} \geq 3$
dấu = khi x=y=z=1