Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$ and $a+b+c+abc=4.$ Prove that: $$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3a}}\geq\frac{a+b+c}{2}.$$
The equality occurs for $a=b=c=1$.
Also, for $c=0$ we obtain: $$\frac{a}{\sqrt{a+3b}}+\sqrt{b}\geq\frac{a+b}{2},$$ which for $b=0$ and $a=4$ gives an equality.
The following Holder saves the cases of the equality occurring.
$$\left(\sum_{cyc}\frac{a}{\sqrt{a+3b}}\right)^2\sum_{cyc}a(a+3b)\geq(a+b+c)^3.$$ Thus, it's enough to prove that: $$4(a+b+c)\geq\sum_{cyc}(a^2+3ab),$$ which would be easy to prove by $uvw$, but it's wrong!
Also, by C-S $$\sum_{cyc}\frac{a}{\sqrt{a+3b}}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}a\sqrt{a+3b}}$$ and it's enough to prove that $$2(a+b+c)\geq\sum_{cyc}a\sqrt{a+3b},$$ which is true for $(a,b,c)=(4,0,0),$ but it's wrong for $a=b=2$ and $c=0$.
I checked that the starting inequality is true for $b=a$ and $c=\frac{4-2a}{1+a^2},$ where $0<a\leq2.$
Thank you!
Note that $\frac{a}{\sqrt{a+3b}} = \frac{4a}{2\sqrt{(a+3b)\cdot 4}} \ge \frac{4a}{4 + a + 3b}$ (AM-GM). It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{4a}{4 + a + 3b}\ge \frac{a+b+c}{2}.$$
By using $c = \frac{4-a-b}{ab + 1}$, after clearing the denominators, it suffices to prove that $f(a, b)\ge 0$ for $a, b \ge 0$ with $a + b \le 4$, where \begin{align} f(a, b) &= 3 a^4 b^3+9 a^3 b^4+12 a^4 b^2+16 a^3 b^3+12 a^2 b^4-9 a^4 b-15 a^3 b^2\\ &\quad +3 a^2 b^3-3 a b^4+44 a^3 b-84 a^2 b^2+20 a b^3-6 a^3+61 a^2 b\\ &\quad +79 a b^2+6 b^3+20 a^2-304 a b-52 b^2+16 a+112 b+60. \end{align} We have $$f(a,b) = \frac{1}{48}g(a,b) + \frac{4-a-b}{48}h(a, b)$$ where \begin{align} g(a, b) &= 144 a^4 b^3+432 a^3 b^4+576 a^4 b^2+768 a^3 b^3+576 a^2 b^4-432 a^4 b\\ &\quad -720 a^3 b^2+144 a^2 b^3-144 a b^4+332 a^4+3513 a^3 b-1901 a^2 b^2\\ &\quad +2285 a b^3+263 b^4-2532 a^3-5512 a^2 b-4732 a b^2-1792 b^3 \\ &\quad +5508 a^2+606 a b+2938 b^2-2508 a+348 b+1840, \\ h(a,b) &= 332 a^3+1069 a^2 b+1062 a b^2+263 b^3-916 a^2-3248 a b\\ &\quad -1028 b^2+884 a+1322 b+260. \end{align} We may use the Buffalo Way to prove that both $g(a,b)\ge 0$ and $h(a,b)\ge 0$ for $a, b \ge 0$. The desired result follows.
We are done.