Let $a,b,c>0$,with $a+b+c=3$,Prove that
$$\sqrt{\frac{a^{3}}{a^{2}+3b^{2}}}+\sqrt{\frac{b^{3}}{b^{2}+3c^{2}}}+\sqrt{\frac{c^{3}}{c^{2}+3a^{2}}}\geq \frac{3}{2}$$

Proof:
$$\begin{align*} \sum_{cyc}{\sqrt{\frac{a^{3}}{a^{2}+3b^{2}}}}&=6\sum_{cyc}{\frac{a^{2}}{\sqrt{4a(a+b+c)\cdot3(a^2+3b^2)}}}\\ &\geq 12\sum_{cyc}{\frac{a^2}{4a(a+b+c)+3(a^2+3b^2)}}\\ &=12\sum_{cyc}{\frac{a^{2}}{7a^{2}+9b^2+4ab+4ca}}  \end{align*}$$
Now,Using the Awesome CYH technology,we have
$$\begin{align*} &\left(\sum_{cyc}{\frac{a^{2}}{7a^{2}+9b^2+4ab+4ca}}\right)\left[\sum_{cyc}{(2a+c)^2(7a^2+9b^2+4ab+4ca)}\right]\\ &\geq \left[\sum_{cyc}{a(c+2a)}\right]^{2}\\ &=\left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2  \end{align*}$$
Thus,it's suffice to check
$$8\left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2\geq \sum_{cyc}{(2a+c)^2(7a^2+9b^2+4ab+4ca)}$$
Or
$$\sum_{cyc}{a^4}+\sum_{cyc}{a^2b^2}+3\sum_{cyc}{a^3b}-3\sum_{cyc}{ab^3}-2abc\sum_{cyc}{a}\geq 0$$
Which is not a problem for us now,so we are done!
$\square$