Ta có \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\)Đặt \(t=x^2-5x+5\), khi đó, ta có VT = \(\left(t-1\right)\left(t+1\right)=t^2-1\ge-1\) (đpcm)
Dấu = xảy ra <=> \(x\in\left\{\dfrac{5+\sqrt{5}}{2},\dfrac{5-\sqrt{5}}{2}\right\}\)