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. 
 A: Hello Michael Rozenberg . I start my proof with a classical substitution .
Put:
$A=a$;
$AB=b$;
$AC=c$;
$AD=d$
We get :
$$\frac{1}{\sqrt{1+3B}}+\frac{B}{\sqrt{B+3C}}+\frac{C}{\sqrt{C+3D}}+\frac{D}{\sqrt{D+3}}\geq \sqrt{1+B+C+D}$$
Make a new substitution like this :
$x=B$;
$C=xh$;
$D=x\beta$
We get :
$$\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\frac{x\beta}{\sqrt{x\beta+3}}\geq \sqrt{1+x(1+h+\beta)}$$
We can rewriting the inequality like this :
$$\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\sqrt{x\beta+3}+\frac{-3}{\sqrt{x\beta+3}}-\sqrt{1+x(1+h+\beta)}\geq 0$$
Now we introduce the following function :
$$f(x)=\frac{1}{\sqrt{1+3x}}+\frac{\sqrt{x}}{\sqrt{1+3h}}+\frac{\sqrt{x}h}{\sqrt{h+3\beta}}+\sqrt{x\beta+3}+\frac{-3}{\sqrt{x\beta+3}}-\sqrt{1+x(1+h+\beta)}$$
The idea is to integrate the function $f$ we get :
$$F(x)=\frac{2}{3}\sqrt{1+3x}+\frac{2}{3}(x)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}}+\frac{2}{3\beta}(x\beta+3)^{\frac{3}{2}}-\frac{6}{\beta}\sqrt{x\beta+3}-\frac{2}{3(1+h+\beta)}(1+x(1+h+\beta))^{\frac{3}{2}}$$
Now the idea is to note that the function $f$ is positiv if the function $F$ is increasing .
Namely when we have :
$F(X)\leq F(Y)$ with $X\leq Y$
Here we take $X=x-\epsilon$ and $Y=y+\epsilon$
We cut the function $F$ in others to a question of space 
We have :
$g(x-\epsilon)-g(x+\epsilon)=\frac{2}{3}\sqrt{1+3(x-\epsilon)}-\frac{2}{3}\sqrt{1+3(x+\epsilon)}$
$u(x-\epsilon)-u(x+\epsilon)=\frac{2}{3}(x-\epsilon)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x-\epsilon)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}}-(\frac{2}{3}(x+\epsilon)^{\frac{3}{2}}\frac{1}{\sqrt{1+3h}}+\frac{2}{3}(x+\epsilon)^{\frac{3}{2}}\frac{h}{\sqrt{h+3\beta}})$
$v(x-\epsilon)-v(x+\epsilon)=\frac{2}{3\beta}((x-\epsilon)\beta+3)^{\frac{3}{2}}-\frac{2}{3\beta}((x+\epsilon)\beta+3)^{\frac{3}{2}}$
$q(x+\epsilon)-q(x-\epsilon)=\frac{6}{\beta}\sqrt{(x+\epsilon)\beta+3}+\frac{2}{3(1+h+\beta)}(1+(x+\epsilon)(1+h+\beta))^{\frac{3}{2}}
-(\frac{6}{\beta}\sqrt{(x-\epsilon)\beta+3}+\frac{2}{3(1+h+\beta)}(1+(x-\epsilon)(1+h+\beta))^{\frac{3}{2}}
)$
Now we have the following elementary inequalities with $n\geq x^6$:
$g(x-\epsilon)-g(x+\epsilon)\leq -\frac{2}{3}\frac{\epsilon}{n} x^6$ 
$u(x-\epsilon)-u(x+\epsilon)\leq -\frac{2\epsilon}{3n}x^6(\frac{1}{\sqrt{1+3h}}+\frac{h}{\sqrt{h+3\beta}})$ 
$v(x-\epsilon)-v(x+\epsilon)\leq -\frac{2\epsilon}{3n}x^6$
$q(x+\epsilon)-q(x-\epsilon)\leq \frac{\epsilon}{n}(\frac{2}{3})x^6$
So you have just to add the elementary inequalities to get the result (and derivate)
If you have questions tell me .
A: 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.
