$a+b+c=3\Rightarrow\sum\limits_{cyc}\frac{a}{b\sqrt{c^2+3}}\geq\frac{a^2+b^2+c^2}{2}$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove that:
$$\frac{a}{b\sqrt{c^2+3}}+\frac{b}{c\sqrt{a^2+3}}+\frac{c}{a\sqrt{b^2+3}}\geq\frac{a^2+b^2+c^2}{2}$$
I tried C-S, AM-GM, Holder and more, but without success. 
Thank you!
 A: The Buffalo Way works although it is an ugly solution.
Squaring both sides of the inequality, we need to prove that
$$\sum_{\mathrm{cyc}} \frac{a^2}{b^2(c^2+3)} + \sum_{\mathrm{cyc}} \frac{2a}{c\sqrt{(c^2+3)(a^2+3)}}\ge \frac{(a^2+b^2+c^2)^2}{4}.$$
Using the Cauchy-Bunyakovsky-Schwarz inequality to get $\sqrt{(c^2+3)(a^2+3)} \ge ca + 3$, it suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{a^2}{b^2(c^2+3)} + \sum_{\mathrm{cyc}} \frac{2a(ca + 3)}{c(c^2+3)(a^2+3)}\ge \frac{(a^2+b^2+c^2)^2}{4}.$$
After homogenization, we need to prove that
$$\sum_{\mathrm{cyc}} \frac{a^2}{b^2(c^2+(a+b+c)^2/3)} + \sum_{\mathrm{cyc}} \frac{2a(ca + (a+b+c)^2/3)}{c(c^2+(a+b+c)^2/3)(a^2+(a+b+c)^2/3)}\ge \frac{(a^2+b^2+c^2)^2}{4(a+b+c)^6/3^6}$$
or
$f(a,b,c)\ge 0$ where $f(a,b,c)$ is a homogeneous polynomial of degree $16$. 
We use the Buffalo Way. WLOG, assume that $c = \min(a,b,c)$.
If $c = 0$, $f(a,b,0)$ is a polynomial with non-negative coefficients. True.
If $c > 0$ and $c \le a \le b$, let $c = 1, \ a = 1+s, \ b= 1+s + t$. 
$f(1+s, 1+s+t, 1)$ is a polynomial with non-negative coefficients. True.
If $c > 0$ and $c \le b\le a$, let $c=1, \ b = 1+s, \ a = 1+s+t$.
$f(1+s+t, 1+s, 1)$ is a polynomial with non-negative coefficients. True.
We are done.
A: Your inequality is equivalent to :
$$\sum_{cyc}\frac{a}{b\sqrt{c^2+\frac{(a+b+c)^2}{3}}}\geq\frac{a^2+b^2+c^2}{(a+b+c)^3}\frac{27}{2}$$
We begin with a first substitution we put :
$\frac{b}{2}-a=-u$
$\frac{b}{2}+a=v$
$c+a=w$
The initial inequality become :
$$\sum_{cyc}\frac{\frac{v+u}{2}}{(v-u)\sqrt{(\frac{2w-u-v}{2})^2+\frac{(v-u+w)^2}{3}}}\geq \frac{(\frac{v+u}{2})^2+(v-u)^2+(\frac{2w-u-v}{2})^2}{(v-u+w)^3}\frac{27}{2}$$
So we start here with a second substitution we put :
$-u=\frac{p}{\sqrt{-r^2+q^2-p^2}}$
$v=\frac{r}{\sqrt{-r^2+q^2-p^2}}$
$w=\frac{q}{\sqrt{-r^2+q^2-p^2}}$
We get this :
$p+r+q=3\sqrt{-r^2+q^2-p^2}$
The initial inequality become 
$$\sum_{cyc}\frac{\frac{-p+r}{2}}{(r+p)\sqrt{(\frac{2q+p-r}{2})^2+\frac{(r+p+q)^2}{3}}}\geq \frac{(\frac{-p+r}{2})^2+(r+p)^2+(\frac{2q-r+p}{2})^2}{(r+p+q)^3}\frac{27}{2}$$
We make a last substitution :
$r=R$
$p=RP$
$q=RQ$
So the initial condition become :
$P+1+Q=3\sqrt{-1+Q^2-P^2}$
Wich is equivalent to :
$Q=\frac{3}{8}\sqrt{(9P^2+2P+9)}+\frac{(1+P)}{8}$
The initial inequality become :
$$\sum_{cyc}\frac{\frac{-P+1}{2}}{(1+P)\sqrt{(\frac{2Q-1+P}{2})^2+\frac{(1+P+Q)^2}{3}}}\geq \frac{(\frac{-P+1}{2})^2+(1+P)^2+(\frac{2Q-1+P}{2})^2}{(1+P+Q)^3}\frac{27}{2}$$
So if you replace the value of $Q$ the inequality is just with the variable $P$
So you have many ways to treat it .
Edit :
We treat the case where $b>2a$ and $a\leq c$
It's easy to see that we have :
$$\sum_{cyc}\frac{a}{b\sqrt{c^2+3}}\geq 3(0.75-a)+0.5\frac{1}{\sqrt{(3-3a)^2+3}}+\frac{2a}{(3-3a)\sqrt{a^2+3}}+\frac{3-3a}{a\sqrt{4a^2+3}}\geq  3(0.75-a)+\frac{a^2+4a^2+(3-3a)^2}{2}\geq\frac{a^2+b^2+c^2}{2}$$
So it's conclude the proof .
