Prove that $\sum\limits_{cyc}\frac{a^2-bd}{b+2c+d}\geq0$ 
Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
  $$\frac{a^2-bd}{b+2c+d}+\frac{b^2-ca}{c+2d+a}+\frac{c^2-db}{d+2a+b}+\frac{d^2-ac}{a+2b+c}\geq0$$

This inequality is a similar to the following inequality of three variables.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2-bc}{b+c+2a}+\frac{b^2-ca}{c+a+2b}+\frac{c^2-ab}{a+b+2c}\geq0,$$
which we can prove by the following reasoning.
$$\sum\limits_{cyc}\frac{a^2-bc}{b+c+2a}=\frac{1}{2}\sum\limits_{cyc}\frac{(a-b)(a+c)-(c-a)(a+b)}{b+c+2a}=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a+c}{b+c+2a}-\frac{b+c}{c+a+2b}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{(b+c+2a)(c+a+2b)}\geq0,$$
but this idea does not help for the starting inequality.
We can make the following. By Holder
$$\sum_{cyc}\frac{a^2}{b+2c+d}=\sum_{cyc}\frac{a^3}{ab+2ac+ad}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(ab+2ac+ad)}=\frac{(a+b+c+d)^3}{8\sum\limits_{cyc}(ab+ac)}$$
and
$$\sum_{cyc}\frac{bd}{b+2c+d}=2(a+b+c+d)\left(\frac{bd}{(b+2c+d)(d+2a+b)}+\frac{ac}{(a+2b+c)(c+2d+a)}\right).$$
Thus, it remains to prove that
$$\frac{(a+b+c+d)^2}{16\sum\limits_{cyc}(ab+ac)}\geq\frac{bd}{(b+2c+d)(d+2a+b)}+\frac{ac}{(a+2b+c)(c+2d+a)}$$
and I don't see, what is the rest.
 A: The following quantity is clearly positive
\begin{eqnarray*}
((2(d^3+b^3)+8bd(b+d))(b-d)^2+(2(a^3+c^3)+8ca(c+a))(c-a)^2)+
((a+c)(5(d^2+b^2)(b-d)^2+12bd(b-d)^2)+(b+d)(5(a^2+c^2)(a-c)^2+12ac(a-c)^2))+
2((b^3+d^3)(a-c)^2+(a^3+c^3)(b-d)^2)+
14(bd(a+c)(a-c)^2+ac(b+d)(b-d)^2)
\end{eqnarray*}
After doing some algebra, this is the same as
\begin{eqnarray*}
(a^2-bd)(c+2d+a)(d+2a+b)(a+2b+c)+(b^2-ca)(d+2a+b)(a+2b+c)(b+2c+d)+(c^2-db)(a+2b+c)(b+2c+d)(c+2d+a)+(d^2-ac)(b+2c+d)(c+2d+a)(d+2a+b) \geq 0
\end{eqnarray*}
Now divide by $(c+2d+a)(d+2a+b)(a+2b+c)(b+2c+d)$ and the result follows.
(a^2-b*d)*(c+2*d+a)*(d+2*a+b)*(a+2*b+c)+(b^2-c*a)*(d+2*a+b)*(a+2*b+c)*(b+2*c+d)+(c^2-d*b)*(a+2*b+c)*(b+2*c+d)*(c+2*d+a)+(d^2-a*c)*(b+2*c+d)*(c+2*d+a)*(d+2*a+b)-(((2*(d^3+b^3)+8*b*d*(b+d))*(b-d)^2+(2*(a^3+c^3)+8*c*a*(c+a))*(c-a)^2)+((a+c)*(5*(d^2+b^2)*(b-d)^2+12*b*d*(b-d)^2)+(b+d)*(5*(a^2+c^2)*(a-c)^2+12*a*c*(a-c)^2))+2*((b^3+d^3)*(a-c)^2+(a^3+c^3)*(b-d)^2)+14*(b*d*(a+c)*(a-c)^2+a*c*(b+d)*(b-d)^2));

The above computer algebra can be copied & pasted into reduce & serves as justification for the above claim.
The above solution does not use any well known theorems or tricks. I am sure that more elegant solutions exist & We would be interested to see them.  
A: Update
By chance, I saw a nice proof as follows.
\begin{align}
\sum_{\mathrm{cyc}} \frac{a^2-bd}{b+2c+d}
&\ge \sum_{\mathrm{cyc}} \frac{a^2-\frac{(b+d)^2}{4}}{b+2c+d}\\
&= \sum_{\mathrm{cyc}} \Big(\frac{a^2-\frac{(b+d)^2}{4}}{b+2c+d} + \frac{b+d-2c}{4}\Big)\\
&= \sum_{\mathrm{cyc}} \frac{a^2-c^2}{b+2c+d}\\
&= \Big(\frac{a^2-c^2}{b+2c+d} + \frac{c^2-a^2}{d+2a+b}\Big) + \Big(\frac{b^2-d^2}{c+2d+a} + \frac{d^2-b^2}{a+2b+c}\Big)\\
&= \frac{2(a+c)(a-c)^2}{(b+2c+d)(d+2a+b)} + \frac{2(b+d)(b-d)^2}{(c+2d+a)(a+2b+c)}\\
&\ge 0.
\end{align}
Previously written
Donald Splutterwit gave a nice SOS (Sum of Squares) solution.
Actually, the Buffalo Way works.
After clearing the denominators, we need to prove that $f(a,b,c, d)\ge 0$ where $f(a,b,c,d)$ is
a homogeneous polynomial of degree four.
WLOG, assume that $d = \min(a,b,c,d)$. Let $c = d+s, \ b=d+t, \ a = d+r; \ s, t, r\ge 0$.
We have $f(d+r, d+t, d+s, d) = Ad^2 + Bd + C$ where
\begin{align}
A &= 24 r^2-48 r s+24 s^2+24 t^2, \\
B &= 16 r^3-16 r^2 s+8 r^2 t-16 r s^2-16 r s t+8 r t^2+16 s^3+8 s^2 t+8 s t^2+16 t^3, \\
C &= 2 r^4+2 r^3 s+3 r^3 t-8 r^2 s^2-3 r^2 s t-r^2 t^2+2 r s^3-3 r s^2 t \\
&\quad +4 r s t^2+3 r t^3+2 s^4+3 s^3 t-s^2 t^2+3 s t^3+2 t^4.
\end{align}
It suffices to prove that $A, B, C\ge 0$. Clearly $A\ge 0$. It is easy to prove that $B\ge 0$ using discriminant. The proof of $C\ge 0$ may be a little harder. Omitted here. However I also verified it by Mathematica Resolve.
Maybe someone can find a nice proof of $B, C\ge 0$.
