Find all quadruplets (x, y, z, w) of positive real numbers that satisfy the following system:
$\begin{cases}
\frac{xyz+1}{x+1}= \frac{yzw+1}{y+1}= \frac{zwx+1}{z+1}= \frac{wxy+1}{w+1}\\
x+y+z+w= 48
\end{cases}$
Given the system of equations:
\[
\begin{cases}
\frac{xyz+1}{x+1}= \frac{yzw+1}{y+1}= \frac{zwx+1}{z+1}= \frac{wxy+1}{w+1} \\
x+y+z+w= 48
\end{cases}
\]

Let \( u > 0 \) be the common value of the first equation. Therefore, we have:
\[
\frac{xyz+1}{x+1} = u, \quad \frac{yzw+1}{y+1} = u, \quad \frac{zwx+1}{z+1} = u, \quad \frac{wxy+1}{w+1} = u
\]

This implies:
\[
xyz + 1 = u(x + 1), \quad yzw + 1 = u(y + 1), \quad zwx + 1 = u(z + 1), \quad wxy + 1 = u(w + 1)
\]

Subtracting each equation from the previous, we get:
\[
yz(x - w) = u(x - y), \quad zw(y - x) = u(y - z), \quad wx(z - y) = u(z - w), \quad xy(w - z) = u(w - x)
\]

If two of \( x, y, z, w \) are equal, then all are equal. Suppose \( x = y = z = w \). Then from the second equation:
\[
x + y + z + w = 48 \implies 4x = 48 \implies x = 12
\]

Thus, the solution in this case is:
\[
\boxed{x = y = z = w = 12}
\]

Now, consider the case where \( x, y, z, w \) are pairwise different. Multiplying the four equations, we get:
\[
(xyz + 1)(yzw + 1)(zwx + 1)(wxy + 1) = u^4 (x + 1)(y + 1)(z + 1)(w + 1)
\]

Since \( xyzw = u^2 \), the original equations become:
\[
\frac{u^2}{w} + 1 = u(x + 1), \quad \frac{u^2}{x} + 1 = u(y + 1), \quad \frac{u^2}{y} + 1 = u(z + 1), \quad \frac{u^2}{z} + 1 = u(w + 1)
\]

Subtracting each equation from the previous, we get:
\[
xy(z - y) = u(x - y), \quad yz(w - z) = u(y - z), \quad zw(x - w) = u(z - w), \quad wx(y - x) = u(w - x)
\]

Comparing the first elements of lines (1) and (3), we get:
\[
yz(x - w) = xy(z - y) \implies zw = xy \implies zw = xy = u
\]

Similarly, we get:
\[
wx = yz = u, \quad xy = zw = u, \quad yz = wx = u
\]

This implies that \( x, y, z, w \) must be equal, which contradicts the assumption that they are pairwise different. Therefore, no other solutions exist.

\(\blacksquare\)