$$\frac{a}{\sqrt{5-4bc}} + \frac{b}{\sqrt{5-4ac}} + \frac{c}{\sqrt{5-4ab}} \ge 1$$ With the condition $a+b+c=3$
I am thinking about some type of Cauchy-Schwarz inequality here, but not quite sure how to apply it. I believe 5 could be seperated to $a + b + c + 2 = 5$ and see if it helps under the square root but still , I tried that and didn't really find anything helpful. Probably some other transformation takes place ?
Then this came to my mind $$a + b + c = 3 => \frac{a}{3} + \frac{b}{3} + \frac{c}{3} = 1$$
Thus I could try to prove that the first fractions is $\ge \frac{a}{3}$ and so on , then sum side by side and get $LHS \ge \frac{a}{3} + \frac{b}{3} + \frac{c}{3} =1$
Currently I am experimenting with this idea, but I can see it doesn't bring me far. Any help is appreciated.
Quick update, I need to show that $bc \le 1$ and I guess it's over.
Here is one approach that is possibly overkill and provides a sharper bound. First, rewrite each fraction so that the product terms become the symmetric term $P=abc$.
$$S=\sum_{\mathrm{cyc}}\frac{a}{\sqrt{5-4bc}}=\sum_{\mathrm{cyc}}\frac{a^{3/2}}{\sqrt{5a-4P}}$$
Note that the function $\displaystyle F(x)=\frac{x^{3/2}}{\sqrt{5x-4P}}$ is convex (I will leave that as an exercise for you), so we can use Jensen's inequality, which states that $a_1F(x_1)+\ldots+a_nF(x_n)\ge F(a_1x_1+\ldots+a_nx_n)$ where $F$ is convex, all $a_i\ge 0$, and $\sum a_i=1$. Let us take $a_i=1/3$.
Hence,
$$S= 3 \sum_{\mathrm{cyc}}\frac13F(a)\ge 3\cdot F\left(\frac13a+\frac13b+\frac13c\right)=\frac{3}{\sqrt{5-4P}}$$
Observe from the binomial series that $\displaystyle \frac{1}{\sqrt{1-x}}\ge 1+\frac{x}{2}$ everywhere the function is defined. Hence, $S\ge \frac{3}{\sqrt{5}}\left(1+\frac{2}{5}P\right)\ge \frac{3}{\sqrt{5}}=1.34$.
As noted in the comments below this answer, your problem is missing the restriction that $a,b,c\ge 0$, without which it is not necessarily true, since the expression would otherwise be zero at $(a,b,c)=(3,\pm0.409434,\mp0.409434)$.