$a,b,c,d$ are fixed positive real numbers. Find the maximum value of the function $f: \mathbb{R^{+}}_{0}  \rightarrow \mathbb{R}$ $f(x)=\frac{a+bx}{b+cx}+\frac{b+cx}{c+dx}+\frac{c+dx}{d+ax}+\frac{d+ax}{a+bx}, x \geq 0$
1. Let \( a_1 = \frac{a}{b} \), \( a_2 = \frac{b}{c} \), \( a_3 = \frac{c}{d} \), and \( a_4 = \frac{d}{a} \). We claim that the maximum value of \( f(x) \) on the nonnegative reals is \( a_1 + a_2 + a_3 + a_4 \). This is attained at \( x = 0 \).

2. To show that this is the maximum value, we start by rewriting the function \( f(x) \):
   \[
   f(x) = \frac{a+bx}{b+cx} + \frac{b+cx}{c+dx} + \frac{c+dx}{d+ax} + \frac{d+ax}{a+bx}
   \]

3. Consider the first term \( \frac{a+bx}{b+cx} \):
   \[
   \frac{a+bx}{b+cx} = \frac{b \cdot \frac{a}{b} + bx}{b + cx} = \frac{a_1 + x}{1 + \frac{cx}{b}} = \frac{a_1 + x}{1 + \frac{x}{a_2}}
   \]
   Simplifying further:
   \[
   \frac{a_1 + x}{1 + \frac{x}{a_2}} = \frac{a_1a_2 + xa_2}{a_2 + x} = a_1 + \frac{x(a_2 - a_1)}{a_2 + x}
   \]

4. Applying the same transformation to the other terms, we get:
   \[
   f(x) = \left( a_1 + \frac{x(a_2 - a_1)}{a_2 + x} \right) + \left( a_2 + \frac{x(a_3 - a_2)}{a_3 + x} \right) + \left( a_3 + \frac{x(a_4 - a_3)}{a_4 + x} \right) + \left( a_4 + \frac{x(a_1 - a_4)}{a_1 + x} \right)
   \]

5. To show that \( f(x) \leq a_1 + a_2 + a_3 + a_4 \), we need to show:
   \[
   \sum_{cyc} \frac{x(a_2 - a_1)}{a_2 + x} \leq 0
   \]

6. Observe that:
   \[
   \sum_{cyc} \frac{a_2 - a_1}{a_2 + x} = 4 - \sum_{cyc} \frac{a_1 + x}{a_2 + x}
   \]

7. By the AM-GM inequality, we have:
   \[
   \sum_{cyc} \frac{a_1 + x}{a_2 + x} \geq 4
   \]

8. Therefore:
   \[
   4 - \sum_{cyc} \frac{a_1 + x}{a_2 + x} \leq 4 - 4 = 0
   \]

9. This implies:
   \[
   \sum_{cyc} \frac{a_2 - a_1}{a_2 + x} \leq 0
   \]

10. Hence, the maximum value of \( f(x) \) is \( a_1 + a_2 + a_3 + a_4 \), which is attained at \( x = 0 \).

The final answer is \( \boxed{ a_1 + a_2 + a_3 + a_4 } \).