Prove that if $d=0$, then $a b c \leqslant \frac{1}{27}$, the inequality is obviously true. If $a, b, c, d>0$, we only need to prove:

Let
$$f(a, b, c, d)=\sum_{c y c} \frac{1}{a}-\frac{1}{27} \frac{1}{a b c d} \leqslant \frac{176}{27}$$

Then
$$a \leqslant \frac{1}{4} \leqslant b, a^{\prime}=\frac{1}{4}, b^{\prime}=a+b-\frac{1}{4}$$
$$a+b=a^{\prime}+b^{\prime}, a^{\prime} b^{\prime} \geqslant a b .$$

Thus
$$\begin{aligned} & f(a, b, c, d)-f\left(a^{\prime}, b^{\prime}, c, d\right) \\ = & \left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a}-\frac{1}{b^{\prime}}\right)-\frac{1}{27} \cdot \frac{1}{c d} \cdot\left(\frac{1}{a b}-\frac{1}{a^{\prime} b^{\prime}}\right) \\ = & \frac{a^{\prime} b^{\prime}-a b}{a b a^{\prime} b^{\prime}}(a+b)\left(1-\frac{1}{27} \frac{1}{c d(a+b)}\right) \\ \leqslant & \frac{a^{\prime} b^{\prime}-a b}{a b a^{\prime} b^{\prime}}(a+b)\left[1-\frac{1}{27} \cdot\left(\frac{a+b+c+d}{3}\right)^{-3}\right] \\ = & 0 \end{aligned}$$

Therefore
$$f(a, b, c, d) \leqslant f\left(a^{\prime}, b^{\prime}, c, d\right)$$

After at most two more smoothing transformations, we get
$$\begin{aligned} f(a, b, c, d) & \leqslant f\left(\frac{1}{4}, a+b-\frac{1}{4}, c, d\right) \\ & \leqslant f\left(\frac{1}{4}, \frac{1}{4}, a+b+c-\frac{1}{2}, d\right) \\ & \leqslant f\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)=\frac{176}{27} \end{aligned}$$

Thus the original inequality holds.