<s>[INST] $a+b+c=3\Rightarrow\sum\limits_{cyc}\frac{a}{b\sqrt{c^2+3}}\geq\frac{a^2+b^2+c^2}{2}$ 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!
 [/INST] 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.
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