If $a+b+c=3$ so $\sum\limits_{cyc}\frac{a}{\sqrt{b^2+c^2}}\geq\sum\limits_{cyc}\frac{1}{\sqrt{a^2+bc}}$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove that:
$$\frac{a}{\sqrt{b^2+c^2}}+\frac{b}{\sqrt{a^2+c^2}}+\frac{c}{\sqrt{a^2+b^2}}\geq\frac{1}{\sqrt{a^2+bc}}+\frac{1}{\sqrt{b^2+ac}}+\frac{1}{\sqrt{c^2+ab}}$$
I tried to use C-S, Holder, SOS, Rearrangement and more, but without some success.

The Buffalo Way works although it is an ugly solution.
Squaring both sides of the inequality, we need to prove that
$$\sum_{\mathrm{cyc}} \frac{a^2}{b^2+c^2} + \sum_{\mathrm{cyc}} \frac{2ab}{\sqrt{b^2+c^2}\sqrt{c^2+a^2}}
\ge \sum_{\mathrm{cyc}} \frac{1}{a^2+bc} + \sum_{\mathrm{cyc}} \frac{2}{\sqrt{a^2+bc}\sqrt{b^2+ca}}.$$
Using AM-GM and GM-HM, we have $\sqrt{b^2+c^2}\sqrt{c^2+a^2} \le \frac{b^2+c^2 + c^2 + a^2}{2}$
and $\sqrt{a^2+bc}\sqrt{b^2+ca} \ge \frac{2(a^2+bc)(b^2+ca)}{a^2+bc + b^2+ca}$. Thus, it suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{a^2}{b^2+c^2} + \sum_{\mathrm{cyc}} \frac{4ab}{b^2+c^2 + c^2 + a^2}
\ge \sum_{\mathrm{cyc}} \frac{1}{a^2+bc} + \sum_{\mathrm{cyc}} \frac{a^2+bc + b^2+ca}{(a^2+bc)(b^2+ca)}.$$
After homogenization, we need to prove that
$$\sum_{\mathrm{cyc}} \frac{a^2}{b^2+c^2} + \sum_{\mathrm{cyc}} \frac{4ab}{b^2+c^2 + c^2 + a^2}
\ge \frac{(a+b+c)^2}{9}\Big(\sum_{\mathrm{cyc}} \frac{1}{a^2+bc} + \sum_{\mathrm{cyc}} \frac{a^2+bc + b^2+ca}{(a^2+bc)(b^2+ca)}\Big)$$
or
$f(a, b, c)\ge 0$ where $f(a,b,c)$ is a homogeneous polynomial with degree $18$.
We use the Buffalo Way. WLOG, assume that $c = \min(a,b,c)$.
If $c = 0$, we have $f(a,b,0) = a^3b^3(2a^2-3ab+2b^2)(2a^2+b^2)(a^2+2b^2)(a^2+b^2)(a-b)^2(a+b)^2$. True.
If $c > 0$ and $c \le a \le b$, let $c = 1, \ a = 1+s, \ b= 1+s + t; \ s,t\ge 0$.
$f(1+s, 1+s+t, 1)$ is a polynomial in $s, t$ with non-negative coefficients. True.
If $c > 0$ and $c \le b\le a$, let $c=1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$.
$f(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. True.
We are done.
