SOLUTION. In this problem, we fix $a+b=m$ and $c+d=n$. Let $x=ab$ and $y=cd$ then
$$ab c + bc d + cd a + da b - \frac{1}{27} - \frac{176}{27} abcd = m y + n x - \frac{1}{27} - \frac{176 xy}{27} = f(x, y)$$
is a linear (convex) function in both $x$ and $y$. It only reaches the maximum at boundary values, namely
$$\max f(x, y) = f(\alpha, \beta) ; \alpha \in \left\{0, \frac{m^2}{4}\right\} ; \beta \in \left\{0, \frac{n^2}{4}\right\}$$

If $\alpha = \frac{m^2}{4}$ and $\beta = \frac{n^2}{4}$, we have $a=b, c=d$. In this case, the problem becomes
$$2\left(a^2 c + c^2 a\right) \leq \frac{1}{27} + \frac{176 a^2 c^2}{27}$$
for all non-negative real numbers $a, c$ and $a + c = \frac{1}{2}$. This inequality is equivalent to
$$176 a^2 c^2 + 1 \geq 27 a c$$
which is obviously true since $a c \leq \frac{1}{16}$. The equality holds for $a = b = c = d = \frac{1}{4}$. Otherwise, if $\alpha \neq \frac{m^2}{4}$ and $\beta \neq \frac{n^2}{4}$, we must have $mn = 0$ or $abcd = 0$. WLOG, assume that $d = 0$, the inequality becomes $abc \leq \frac{1}{27}$ if $a + b + c = 1$. This follows immediately from AM-GM inequality and attains equality for $a = b = c = \frac{1}{3}$. We are done.