Help me Tính min: (x+1)(x+2)(x+3)(x+4)

$(x+1)(x+2)(x+3)(x+4)$

$=(x+1)(x+4)(x+2)(x+3)$

$=(x²+5x+4)(x²+5x+6)$

Đặt $x²+5x=t$

$A=(t+4)(t+6)$

$=t²+10t+24$

$=t²+10t+25-1$

$=(t+5)²-1$

Vì $(t+5)²≥0$

$→(t+5)²-1≥-1$

$→$ Dấu “=” xảy ra khi $t+5=0$

$↔x²+5x+5=0$

$↔x²+2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{5}{4}=0$

$↔(x+\dfrac{5}{2})²=\dfrac{5}{4}$

\(\leftrightarrow\left[ \begin{array}{l}x+\dfrac{5}{2}=\dfrac{\sqrt{5}}{2}\\x+\dfrac{5}{2}=-\dfrac{\sqrt{5}}{2}\end{array} \right.\) \(\leftrightarrow\left[ \begin{array}{l}x=\dfrac{\sqrt{5}-5}{2}\\x=\dfrac{-\sqrt{5}-5}{2}\end{array} \right.\)

$→\min A=-1$