Chú ý BĐT $27abc\le(a+b+c)^3$
$a^2(b-c)\le0$$b^2(c-b)=4\frac b2.\frac b2(c-b)\le\frac4{27}(\frac b2+\frac b2+c-b)^3=\frac 4{27}c^3$
$\Rightarrow P\le c^2(1-\frac{23c}{27})=\frac{54^2}{23^2}.\frac{23c}{54}.\frac{23c}{54}(1-\frac{23c}{27})\le\frac{54^2}{23^2.27}(\frac{23}{54}c+\frac{23}{54}c+1-\frac{23}{27}c)^3=\frac{108}{529}$
Dấu bằng xảy ra tại $(a;b;c)=(0;\frac{12}{23};\frac{18}{23})$