Example 1.34 Find the largest positive real number $k$, such that for any positive real numbers $a, b, c, d$, there is always $(a+b+c)\left[3^{4}(a+b+c+d)^{5}+2^{4}(a+b+c+2 d)^{5}\right] \geqslant k a b c d^{3}$
Let $t=a+b+c, u=\frac{t}{d}$, and set $a=b=c=1, d=3$, we can get $k \leqslant 2^{4} \times 3^{7} \times 5$.
To prove
$$\begin{aligned}
(a+b+c)\left[3^{4}(a+b+c+d)^{5}+2^{4}(a+b+c+2 d)^{5}\right] \geqslant \\
2^{4} \times 3^{7} \times 5 a b c d^{3}
\end{aligned}$$

For this, we can prove a stronger conclusion:
$$\begin{aligned}
(a+b+c)\left[3^{4}(a+b+c+d)^{5}+2^{4}(a+b+c+2 d)^{5}\right] \geqslant \\
2^{4} \times 3^{7} \times 5\left(\frac{a+b+c}{3}\right)^{3} d^{3}
\end{aligned}$$

Then the strengthened proposition is transformed into
$$3^{4}(u+1)^{5}+2^{4}(u+2)^{5} \geqslant 2^{4} \times 3^{4} \times 5 u^{2}$$

By the AM - GM inequality
$$\begin{aligned}
3^{4}(u+1)^{5}+2^{4}(u+2)^{5}= & 3^{4}(u+1)^{5}+2^{4}(u+1+1)^{5} \geqslant \\
& 3^{4}\left(2 u^{\frac{1}{2}}\right)^{5}+2^{4}\left(3 u^{\frac{1}{3}}\right)^{5}= \\
& 2^{4} \times 3^{4}\left(2 u^{\frac{5}{2}}+3 u^{\frac{5}{3}}\right)= \\
& 2^{4} \times 3^{4}\left(u^{\frac{5}{2}}+u^{\frac{5}{2}}+u^{\frac{5}{3}}+u^{\frac{5}{3}}+u^{\frac{5}{3}}\right) \geqslant \\
& 2^{4} \times 3^{4} \times 5 u^{2}
\end{aligned}$$

In conclusion, the maximum value of $k$ is $2^{4} \times 3^{7} \times 5$.