\(P\le\frac{2}{a+b+c+1}-\frac{54}{\left(a+b+c+3\right)^3}\)
Đặt \(a+b+c=x>0\)
\(P\le\frac{2}{x+1}-\frac{54}{\left(x+3\right)^3}=\frac{2}{x+1}-\frac{54}{\left(x+3\right)^3}-\frac{1}{4}+\frac{1}{4}\)
\(P\le\frac{1}{4}-\frac{\left(x-3\right)^2\left(x^2+8x+3\right)}{4\left(x+1\right)\left(x+3\right)^3}\le\frac{1}{4}\)
Dấu "=" xảy ra khi \(x=3\) hay \(a=b=c=1\)