Prove that: $\sqrt{\frac{4x^2+y^2}{3x^2+yz}}+\sqrt{\frac{4y^2+z^2}{3y^2+xz}}+\sqrt{\frac{4z^2+x^2}{3z^2+xy}}\ge\frac{3\sqrt{5}}{2}$ 
Let $x$, $y$ and $z$ be positive numbers. Prove that:
$$\sqrt{\frac{4x^2+y^2}{3x^2+yz}}+\sqrt{\frac{4y^2+z^2}{3y^2+xz}}+\sqrt{\frac{4z^2+x^2}{3z^2+xy}}\ge\frac{3\sqrt{5}}{2}.$$

This problem is similar to very many contest problems, but I think it's hard enough.
I tried to use Holder, C-S, AM-GM and more, but without any success.
For example, by Holder $$\sum_{cyc}\sqrt{\frac{4x^2+y^2}{3x^2+yz}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{4x^2+y^2}{3x^2+yz}}\right)^2\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3}{\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3}}\geq$$
$$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(4x^2+y^2)(kx+my+z)\right)^3}{\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3}},$$ but I did not find non-negatives $k$ and $m$, for which the inequality $$4\left(\sum\limits_{cyc}(4x^2+y^2)(kx+my+z)\right)^3\geq45\sum\limits_{cyc}(4x^2+y^2)^2(3x^2+yz)(kx+my+z)^3$$ is true.
Thank you!

(WARNING: this proof is very bad)
First, define two sequences $a_i$ and $b_i$ where $a_1=4x^2+y^2,a_2=4y^2+z^2,a_3=4z^2+x^2$ and $b_1=\frac{1}{3x^2+yz},b_2=\frac{1}{3y^2+xz},b_3=\frac{1}{3z^2+xy}$.
Since we have two three term sequences, we may now use Holder's inequality:
$$\sqrt{4x^2+y^2+4y^2+z^2+4z^2+x^2}=\sqrt{5x^2+5y^2+5z^2}$$
and
$$\sqrt{\frac{1}{3x^2+yz}+\frac{1}{3y^2+xz}+\frac{1}{3z^2+xy}}$$
Thus, by Holder's inequality:
$$\left(\sqrt{5x^2+5y^2+5z^2}\right)\left(\sqrt{\frac{1}{3x^2+yz}+\frac{1}{3y^2+xz}+\frac{1}{3z^2+xy}}\right) \ge \frac{3\sqrt{5}}{2}$$
Since equality holds when the two sequences are proportional, we may set $x=y=z$ and get:
$$\sqrt{\frac{45}{4}}=\frac{3\sqrt{5}}{2}$$
which is the minimum value of this inequality, we may then say that since:
$$\left(\sqrt{5x^2+5y^2+5z^2}\right)\left(\sqrt{\frac{1}{3x^2+yz}+\frac{1}{3y^2+xz}+\frac{1}{3z^2+xy}}\right)=\sqrt{\frac{4x^2+y^2}{3x^2+yz}} + \sqrt{\frac{4y^2+z^2}{3y^2+xz}} + \sqrt{\frac{4z^2+x^2}{3z^2+xy}}$$
This holds to be true.
