We need to prove that
$$\sum_{cyc}\frac{b^2c}{a^2(\sqrt{3(a^2+b^2+c^2)}c+ab)}\geq\frac{3}{4}\sqrt{\frac{3}{a^2+b^2+c^2}}$$
for positives $a$, $b$ and $c$.
$$\sum_{cyc}\frac{b^2c}{a^2(\sqrt{3(a^2+b^2+c^2)}c+ab)}\geq\frac{3}{4}\sqrt{\frac{3}{a^2+b^2+c^2}}$$ or
$$\sum_{cyc}\frac{6b^2c}{a^2(2\sqrt{3(a^2+b^2+c^2)}3c+6ab)}\geq\frac{3}{4}\sqrt{\frac{3}{a^2+b^2+c^2}}.$$
Now, by AM-GM and C-S we obtain
$$\sum_{cyc}\frac{6b^2c}{a^2(2\sqrt{3(a^2+b^2+c^2)}3c+6ab)}\geq\sum_{cyc}\frac{6b^2c}{a^2(3(a^2+b^2+c^2)+9c^2+6ab)}=$$
$$=2\sum_{cyc}\frac{b^2c}{a^2(4c^2+a^2+b^2+2ab)}=2\sum_{cyc}\frac{b^4c^4}{a^2b^2c^2(4c^3+a^2c+b^2c+2abc)}\geq$$
$$\geq\frac{2(a^2b^2+a^2c^2+b^2c^2)^2}{a^2b^2c^2\sum\limits_{cyc}(4a^3+a^2b+a^2c+2abc)}.$$
Id est, it's enough to prove that
$$8(a^2b^2+a^2c^2+b^2c^2)^2\sqrt{\frac{a^2+b^2+c^2}{3}}\geq3 a^2b^2c^2\sum\limits_{cyc}(4a^3+a^2b+a^2c+2abc).$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$9(9v^4-6uw^3)^2\sqrt{3u^2-2v^2}\geq3w^6(4(27u^3-27uv^2+3w^3)+9uv^2-3w^3+6w^3$$ or $f(w^3)\geq0$, where
$$f(w^3)=3(3v^4-2uw^3)^2\sqrt{3u^2-2v^2}-(36u^3-33uv^2+5w^3)w^6,$$
which is decreasing function, which says that it's enough to prove the last inequality
for a maximal value of $w^3$, which happens for equality case of two variables.
Since the last inequality is homogeneous, we can assume $b=c=1$, which gives
$$8(2a^2+1)^2\sqrt{\frac{a^2+2}{3}}\geq3a^2(4(a^3+2)+2(a^2+a+1)+6a)$$
or $g(a)\geq0$, where
$$g(a)=2\ln(2a^2+1)+\frac{1}{2}\ln(a^2+2)-2\ln{a}-\ln(2a^3+a^2+4a+5)+2\ln2-\frac{3}{2}\ln3.$$
But
$$g'(a)=\frac{(a-1)(2a^5+2a^4+29a^3+17a^2+44a+20)}{a(a^2+2)(2a^2+1)(2a^3+a^2+4a+5)},$$
which gives $a_{min}=1$ and since $g(1)=0,$ we are done!