Có $ab+bc+ca\le\frac{(a+b+c)^2}3=3$$\Rightarrow \frac2{3+ab+bc+ca}\le\frac13$$\sqrt[3]\frac{abc}{(1+a)(1+b)(1+c)}\le\frac13(\sum_{}^{}\frac a{a+1})=\frac13(3-\sum_{}^{}\frac1{a+1})\le1-\frac3{a+b+c+3}=\frac12 $$\Rightarrow P\le\frac56$Dấu bằng khi $a=b=c=1$
Có $ab+bc+ca\le\frac{(a+b+c)^2}3=3$$\Rightarrow \frac2{3+ab+bc+ca}\le\frac13$$\sqrt[3]\frac{abc}{(1+a)(1+b)(1+c)}\le\frac13(\sum_{}^{}\frac a{a+1})=1-\sum_{}^{}\frac1{a+1}\le1-\frac3{a+b+c+3}=\frac12 $$\Rightarrow P\le\frac56$Dấu bằng khi $a=b=c=1$
Có $ab+bc+ca\le\frac{(a+b+c)^2}3=3$$\Rightarrow \frac2{3+ab+bc+ca}\le\frac13$$\sqrt[3]\frac{abc}{(1+a)(1+b)(1+c)}\le\frac13(\sum_{}^{}\frac a{a+1})=
\frac1
3(3-\sum_{}^{}\frac1{a+1}
)\le1-\frac3{a+b+c+3}=\frac12 $$\Rightarrow P\le\frac56$Dấu bằng khi $a=b=c=1$