If $a$, $b$, $c$, $d$ are positive reals so $(a+c)(b+d) = 1$, prove the following inequality would be greater than or equal to $\frac {1}{3}$. 
Let $a$, $b$, $c$, $d$ be real positive reals with $(a+c)(b+d) = 1$. Prove that $\frac {a^3}{b + c + d} + \frac {b^3}{a + c + d} + \frac {c^3}{a + b + d} + \frac {d^3}{a + b + c} \geq \frac {1}{3}$.

So I've been trying to solve this problem, and I've been trying to find a way to modify it into using AM-GM. The issue is that the $(a+c)(b+d) = 1$ is really throwing me off, as I haven't dealt with any inequalities that have used that as a condition yet (most other conditions I have seen go along the lines of $abcd = 1$ or something like that), and I'm not sure how exactly to deal with this inequality. Does anyone have any ideas?
 A: By C-S $$\sum_{cyc}\frac{a^3}{b+c+d}=\sum_{cyc}\frac{a^4}{ab+ac+ad}\geq\frac{(a^2+b^2+c^2+d^2)^2}{\sum\limits_{cyc}(ab+ac+ad)}\geq$$
$$\geq\frac{a^2+b^2+c^2+d^2}{\sum\limits_{cyc}(ab+ac+ad)}\geq\frac{1}{3},$$
where the last inequality it's $$\sum_{sym}(a-b)^2\geq0.$$
A: Another way.
Since by AM-GM
$$1=(a+c)(b+d)\leq\left(\frac{a+c+b+d}{2}\right)^2,$$ we obtain $$a+b+c+d\geq2.$$
Now, let $a=kx$, $b=ky$, $c=kz$ and $d=kt$ such that $k>0$ and $x+y+z+t=4$.
Thus, $$k(x+y+z+t)\geq2,$$ which gives $$k\geq\frac{1}{2}.$$
But, $$\sum_{cyc}\frac{a^3}{b+c+d}=\sum_{cyc}\frac{k^2x^3}{y+z+t}\geq\frac{1}{4}\sum_{cyc}\frac{x^3}{y+z+t}$$ and it's enough to prove that
$$\sum_{cyc}\frac{x^3}{y+z+t}\geq\frac{4}{3}$$ or
$$\sum_{cyc}\left(\frac{x^3}{4-x}-\frac{1}{3}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)(3x^2+3x+4)}{4-x}\geq0$$ or
$$\sum_{cyc}\left(\frac{(x-1)(3x^2+3x+4)}{4-x}-\frac{10}{3}(x-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2(9x+28)}{4-x}\geq0$$ and we are done!
A: Also, we can use Holder here:
$$\sum_{cyc}\frac{a^3}{b+c+d}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c+d)}=\frac{(a+b+c+d)^2}{12}\geq\frac{\left(2\sqrt{(a+c)(b+d)}\right)^2}{12}=\frac{1}{3}.$$
