Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{a+3b}}\geq\sqrt{a+b+c+d}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3d}}+\frac{d}{\sqrt{d+3a}}\geq\sqrt{a+b+c+d}$$
I tried Holder, AM-GM and more, but without success. 

WLOG, assume that $d = \min(a,b,c,d)$.
Squaring both sides and using AM-GM, it suffices to prove that
\begin{align}
&\sum_{\mathrm{cyc}} \frac{a^2}{a+3b}
+ \sum_{\mathrm{cyc}} \frac{4ab}{a+3b + b+3c}
+ \frac{4ac}{a+3b + c+3d} + \frac{4bd}{b+3c + d+3a} \\
\ge\ & a + b + c + d.
\end{align}
After clearing the denominators, it suffices to prove that
$f(a, b, c, d)\ge 0$ where $f(a,b,c,d)$ is a homogeneous polynomial of degree $11$.
The Buffalo Way works. Let $c=d+s, \ b = d+t, \ a = d+r; \ s, t, r\ge 0$.
We have $$f(d+r, d+t, d+s, d) = q_9d^9 + q_8d^8 + \cdots + q_1d + q_0.$$
Here $q_9 = 12582912r^2+16777216rs-20971520rt+12582912s^2-20971520st+12582912t^2$, etc.
It suffices to prove that $q_9, q_8, \cdots, q_0 \ge 0$.
I used Mathematica Resolve and the Buffalo Way to verify it. We are done.
However, nice or simple proof for $q_9, q_8, \cdots, q_0 \ge 0$ is expected.
