8.2. If $a, b, c, d$ are positive real numbers such that $a b=2$ and $c d=27$, find the minimum value of the expression $E=(a+1)(b+2)(c+3)(d+4)$.
Solution: We have $(c+3)(d+4)=c d+4 c+3 d+12=39+4 c+3 d$.

From the inequality of means $4 c+3 d \geq 2 \sqrt{4 c \cdot 3 d}=2 \sqrt{12 c d}=2 \sqrt{12 \cdot 27}=36$.

Thus, $(c+3)(d+4) \geq 39+36=75$, with equality if

$$
\left\{\begin{array} { l } 
{ 4 c = 3 d } \\
{ c d = 2 7 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ c = \frac { 3 } { 4 } d } \\
{ \frac { 3 } { 4 } d ^ { 2 } = 2 7 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
c=\frac{9}{2} \\
d=6
\end{array}\right.\right.\right.
$$

From the inequality of means $(a+1)(b+2)=4+2 a+b \geq 2 \sqrt{a} \cdot 2 \sqrt{2 b}=4 \sqrt{2 \cdot a b}=8$, with equality if $a=1, b=2$, which satisfy the condition $a b=2$.

Thus, we obtain $E=(a+1)(b+2)(c+3)(d+4) \geq 8 \cdot 75=600$ and this value is achieved for $a=1, b=2,(a b=2), c=\frac{9}{2}, d=6(c d=27)$.

Therefore, the minimum value of the expression $E$ is 600.