26. Set $$\begin{aligned} f(a, b, c, d) & =a b c+b c d+c d a+d a b-\frac{176}{27} a b c d \\ & =a b(c+d)+c d\left(a+b-\frac{176}{27} a b\right) . \end{aligned}$$ If $a+b-\frac{176}{27} a b \leq 0$, by the arithmetic-geometric inequality we have $f(a, b, c, d) \leq a b(c+d) \leq \frac{1}{27}$. On the other hand, if $a+b-\frac{176}{27} a b>0$, the value of $f$ increases if $c, d$ are replaced by $\frac{c+d}{2}, \frac{c+d}{2}$. Consider now the following fourtuplets: $$\begin{gathered} P_{0}(a, b, c, d), P_{1}\left(a, b, \frac{c+d}{2}, \frac{c+d}{2}\right), P_{2}\left(\frac{a+b}{2}, \frac{a+b}{2}, \frac{c+d}{2}, \frac{c+d}{2}\right), \\ P_{3}\left(\frac{1}{4}, \frac{a+b}{2}, \frac{c+d}{2}, \frac{1}{4}\right), P_{4}\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right) \end{gathered}$$ From the above considerations we deduce that for $i=0,1,2,3$ either $f\left(P_{i}\right) \leq f\left(P_{i+1}\right)$, or directly $f\left(P_{i}\right) \leq 1 / 27$. Since $f\left(P_{4}\right)=1 / 27$, in every case we are led to $$f(a, b, c, d)=f\left(P_{0}\right) \leq \frac{1}{27}$$ Equality occurs only in the cases $(0,1 / 3,1 / 3,1 / 3)$ (with permutations) and ( $1 / 4,1 / 4,1 / 4,1 / 4)$. Remark. Lagrange multipliers also work. On the boundary of the set one of the numbers $a, b, c, d$ is 0 , and the inequality immediately follows, while for an extremum point in the interior, among $a, b, c, d$ there are at most two distinct values, in which case one easily verifies the inequality.