ta có
\(P=\left(\left(x+1\right)\left(x+4\right)\right)\left(\left(x+2\right)\left(x+3\right)\right)+1\)
\(P=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
đặt \(\left(x^2+5x+4\right)=a\)thì \(\left(x^2+5x+6\right)=a+2\) ta có
\(P=a\left(a+2\right)+1\)
\(P=a^2+2a+1\)
\(P=\left(a+1\right)^2\ge0\)