Prove that: $\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\leq\frac{9(a^2+b^2+c^2)}{2(a+b+c)^2}$ Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a^2+3bc}}+\frac{b}{\sqrt{b^2+3ac}}+\frac{c}{\sqrt{c^2+3ab}}\leq\frac{9(a^2+b^2+c^2)}{2(a+b+c)^2}$$
I tried Cauchy-Schwarz:
$$\left(\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\right)^2\leq(a+b+c)\sum_{cyc}\frac{a}{a^2+3bc}.$$
Hence, it remains to prove that
$$\sum_{cyc}\frac{a}{a^2+3bc}\leq\frac{81(a^2+b^2+c^2)}{4(a+b+c)^5},$$
which is wrong for $c\rightarrow0^+$.
Also we can use the following C-S:
$$\left(\sum\limits_{cyc}\frac{a}{\sqrt{a^2+3bc}}\right)^2\leq(1+1+1)\sum_{cyc}\frac{a^2}{a^2+3bc}.$$ 
Thus, it remains to prove that
$$\sum_{cyc}\frac{a^2}{a^2+3bc}\leq\frac{27(a^2+b^2+c^2)^2}{4(a+b+c)^4},$$
which is wrong again: $b=c=1$. 
 A: Using the substitutions $(a, b, c) \to (a^2, b^2, c^2)$, the inequality becomes
$$\sum_{\mathrm{cyc}}\frac{a^2}{\sqrt{a^4+3b^2c^2}} \le \frac{9(a^4+b^4+c^4)}{2(a^2+b^2+c^2)^2}.$$
Using the Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align}
\mathrm{LHS}^2 &= \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2}
+ \sum_{\mathrm{cyc}} \frac{2a^2b^2}{\sqrt{(a^4+3b^2c^2)(b^4+3c^2a^2)}}\\
&\le \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2}
+ \sum_{\mathrm{cyc}} \frac{2a^2b^2}{a^2b^2 + 3abc^2}.
\end{align}
It suffices to prove that
$$\frac{81(a^4+b^4+c^4)^2}{4(a^2+b^2+c^2)^4}\ge \sum_{\mathrm{cyc}} \frac{a^4}{a^4 + 3b^2c^2}
+ \sum_{\mathrm{cyc}} \frac{2a^2b^2}{a^2b^2 + 3abc^2}.$$
After clearing the denominators, it suffices to prove that $f(a, b, c)\ge 0$ where
$f(a,b,c)$ is a homogeneous polynomial of degree $26$.
Due to symmetry and homogeneity, WLOG, assume that $1 = c \le b \le a$.
Let $b = 1 + s, \ a = 1+s + t; \ s,t \ge 0$. $f(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients.
We are done.
