30. Let the substitution $b=a x, c=a y, d=a z$, then by the conditions of the problem, we have $1 \leqslant x \leqslant y \leqslant z$. The inequality in the problem is
$$a^{a x}(a x)^{a y}(a y)^{a z}(a z)^{a} \geqslant(a x)^{a}(a y)^{a x}(a z)^{a y} a^{a z}$$

After canceling $a^{a} a^{a x} a^{a y} a^{a z}$, and taking the $\frac{1}{a}$ power on both sides of the inequality, we get the equivalent inequality $x^{y} y^{z} z \geqslant x y^{x} z^{\prime}$

Let $y=x s, z=x t$, then by $x \leqslant y \leqslant z$, we get $1 \leqslant s \leqslant t$. And since $x \geqslant 1$, we have $y \geqslant s$. Thus, the above inequality becomes
$$x^{n s} y^{n t} x t \geqslant x y^{*}(x t)^{n}$$

After canceling $x^{18} y^{x} x t$, and taking the $\frac{1}{x}$ power on both sides of the inequality, we get the equivalent inequality $y^{t-1} \geqslant$ $t^{\frac{x-1}{x}}$

If $y=1$, then since $x=1, s=\frac{y}{x}=1, y^{t-1}=1=t^{1-1}=1$, the inequality holds. If $t=1$, then since $y^{t-1}=y^{0}=1=1^{\frac{t-1}{x}}=1$, the inequality holds.
If $y>1, t>1$, then taking the $\frac{1}{t-1} \cdot \frac{y}{y-1}(>0)$ power on both sides of the above inequality, we get the equivalent inequality
$$y^{\frac{1}{n-1}} \geqslant t^{\frac{1}{t-1}}(1 \leqslant s \leqslant t, s \leqslant y)$$

We will prove this inequality in two cases.
(1) Suppose $y \geqslant t$. Then when $x>1$, the function $f(x)=x^{\frac{x}{x-1}}$ is a monotonically increasing function. This can be proven by showing its derivative $f^{\prime}(x)>0$.

Indeed,
$$\begin{array}{c} f^{\prime}(x)=\left(e^{\frac{x}{x-1} \ln x}\right)^{\prime}=e^{\frac{x}{x-1} \ln x}\left(\frac{1}{x-1}-\frac{\ln x}{(x-1)^{2}}\right)= \\ x^{\frac{x}{1-1}} \frac{1}{(x-1)^{2}}(x-1-\ln x)>0 \end{array}$$

The last inequality can be derived from the inequality $x-1-\ln x>0$, which is because the function $g(x)=x-$ $1-\ln x$ is an increasing function on $(\Gamma,+\infty)$. This can be derived from $g^{\prime}(x)=1-\frac{1}{x}>0$ and $g(1)=0$.

Since $f(x)=x^{\frac{x}{x-1}}$ is an increasing function on $(\mathrm{F},+\infty)$, we have $y^{\frac{y}{y-1}}=f(y) \geqslant f(t)=t^{\frac{1}{t-i}} \geqslant$ $t^{\frac{1}{1-1}}$.
(2) Suppose $y1$ when the function $f(x)=x^{\frac{1}{x-1}}$ is a monotonically decreasing function. This can be proven by showing its derivative $f^{\prime}(x)<0$.

Indeed, $f^{\prime}(x)=\left(e^{\frac{1}{x-1} \ln x}\right)^{\prime}=e^{\frac{1}{x-1} \ln x}\left(\frac{1}{x(x-1)}-\frac{\ln x}{(x-1)^{2}}\right)=x^{\frac{1}{x-1}} \frac{1}{x(x-1)^{2}}$ $(x-1-x \ln x)<0$.

The last inequality can be derived from the inequality $x-1-x \ln x<0$, which is because the function $g(x)=x-$ $1-x \ln x$ is a decreasing function on $(1,+\infty)$. This can be derived from $g^{\prime}(x)=1-\ln x-1=-\ln x<0$ and $g(1)=0$.

Since $f(x)=x^{\frac{1}{x-1}}$ is a decreasing function on $(1,+\infty)$, we have $y^{\frac{y}{y-1}}=(f(y))^{y} \geqslant$ $(f(t))^{y}=t^{\frac{1}{t-1}} \geqslant t^{\frac{1}{1-1}}$.