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.
 A: The Buffalo Way works although it is an ugly solution.
Squaring both sides of the inequality, we need to prove that
$$\sum_{\mathrm{cyc}} \frac{a^2}{b^2+c^2} + \sum_{\mathrm{cyc}} \frac{2ab}{\sqrt{b^2+c^2}\sqrt{c^2+a^2}}
\ge \sum_{\mathrm{cyc}} \frac{1}{a^2+bc} + \sum_{\mathrm{cyc}} \frac{2}{\sqrt{a^2+bc}\sqrt{b^2+ca}}.$$
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.
A: We begin with a first substitution we put :
$\frac{b}{2}-a=x-2\epsilon<0=-u$
$\frac{b}{2}+a=y+\epsilon=v$
$c+a=z+\epsilon=w$
The initial inequality become :
$$\frac{v-u}{\sqrt{(\frac{v+u}{2})^2+(\frac{2w-v-u}{2})^2}}+\frac{\frac{v+u}{2}}{\sqrt{(v-u)^2+(\frac{2w-v-u}{2})^2}}+\frac{\frac{2w-v-u}{2}}{\sqrt{(v-u)^2+(\frac{u+v}{2})^2}}\geq \frac{1}{\sqrt{(v-u)^2+(\frac{2w-v-u}{2})(\frac{v+u}{2})}}+\frac{1}{\sqrt{(\frac{v+u}{2})^2+(\frac{2w-v-u}{2})(v-u)}}\frac{1}{\sqrt{(\frac{2w-v-u}{2})^2+(v-u)(\frac{v+u}{2})}}$$
We get $w+v-u=3$
Now we make a second substitution :
$-u=\frac{-p}{\sqrt{r^2+q^2-p^2}}$
$v=\frac{r}{\sqrt{r^2+q^2-p^2}}$
$w=\frac{q}{\sqrt{r^2+q^2-p^2}}$
We get this :
$-p+r+q=3\sqrt{r^2+q^2-p^2}$
The initial inequality become 
$$\frac{2\frac{r-p}{2}}{\sqrt{(\frac{r+p}{2})^2+(\frac{2q-r-p}{2})^2}}+\frac{\frac{r+p}{2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{2q-r-p}{2})^2}}+\frac{\frac{2q-r-p}{2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{r+p}{2})^2}}\geq \frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{2q-r-p}{2})(\frac{r+p}{2})}}+\frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(\frac{r+p}{2})^2+(\frac{2q-r-p}{2})(2\frac{r-p}{2})}}\frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(\frac{2q-r-p}{2})^2+(2\frac{r-p}{2})(\frac{r+p}{2})}}$$
We make a last substitution :
$r=R$
$-p=-RP$
$q=RQ$
With $P\geq 1$ if we put $p\geq q \geq r$
So the initial condition become :
$-P+1+Q=3\sqrt{1+Q^2-P^2}$
Wich is equivalent to :
$Q=\frac{3}{8}\sqrt{(9P^2-2P-7)}+\frac{(1-P)}{8}$
The initial inequality become :
$$\frac{2\frac{1-P}{2}}{\sqrt{(\frac{1+P}{2})^2+(\frac{2Q-1-P}{2})^2}}+\frac{\frac{1+P}{2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{2Q-1-P}{2})^2}}+\frac{\frac{2Q-1-P}{2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{1+P}{2})^2}}\geq \frac{\sqrt{1+Q^2-P^2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{2Q-1-P}{2})(\frac{1+P}{2})}}+\frac{\sqrt{1+Q^2-P^2}}{\sqrt{(\frac{1+P}{2})^2+(\frac{2Q-1-P}{2})(2\frac{1-P}{2})}}\frac{\sqrt{1+Q^2-P^2}}{\sqrt{(\frac{2Q-1-P}{2})^2+(2\frac{1-P}{2})(\frac{1+P}{2})}}$$
So to conclude it sufficient to combine the condition with the inequality to obtain an inequality with one variable .
