Inequality in sum of fractions 
Is there a constant $C\ge 0$ such that for any $a,b,c,d>0$ the inequality holds:
  $$
\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge C+\frac{a+b}{b+c}+\frac{b+c}{c+d}+\frac{c+d}{d+a}
$$


My attempt:
$a=b=c=d$ gives $1\ge C$.
Now, take $b=c=d$; this leads to inequality with two variables ($a$ and, let's say, $b$), which for $C=1$ is equivalent to $a^3+2b^3\ge 3ab^2$. This one is true by Cauchy.
The case $a=b$, $c=d$ leads to the same thing.
Is $C=1$ good?

Using scaling of  $a,b,c,d$ you can set $d=1$ without loss of generality. 
This gives the function
$$
f(a,b,c) \equiv \left( \frac{a}{b} + \frac{b}{c} + c + \frac{1}{a} \right) - \left( \frac{a+b}{b+c} + \frac{b+c}{c+1} + \frac{c+1}{1+a}\right)
$$
to minimize.
Taking the derivative $\frac{\partial f(a,b,c)}{\partial b}$ gives a result that is linear in $a$, hence equating to zero gives that for the extremum one gets 
$$
a = \frac{b^2(b^2 + 2 b c -c^3)}{c^2 (1+c) (2 b +c)}
$$
After that it becomes rather messy.
It is of course cheating but a numerical minimisation of the function $f(a,b,c)$ gives that the minimum is obtained for $a \approx 1.193389$, $b \approx 2.201712$, $c \approx 1.494538$ and results in $C \approx 0.810126$.
