Given non-negative real numbers $a,b,c$ satisfying $a^2+b^2+c^2=3.$ Prove that $$\color{black}{\frac{bc}{\sqrt{3a+1}}+\frac{ca}{\sqrt{3b+1}}+\frac{ab}{\sqrt{3c+1}}\le \frac{3}{2}.}$$
Here is my attempts.
Since equality holds at $a=b=c=1$ or $a=b=\sqrt{\dfrac{3}{2}};c=0.$ we can't use normal approach.
For example, it's easy to see that $3=a^2+b^2+c^2\ge \dfrac{(a+b+c)^2}{3} \implies 3\ge a+b+c.$
Thus, $$\sum_{cyc}\frac{bc}{\sqrt{3a+1}}\le \sum_{cyc}\frac{bc}{\sqrt{a(a+b+c)+1}}$$ but $$\sum_{cyc}\frac{bc}{\sqrt{a(a+b+c)+1}}\le \frac{3}{2}$$ is already wrong at $a=b=\sqrt{\dfrac{3}{2}};c=0.$
Remark.
My teacher asasigned this problem to our class as a homework. I post it here to look for help and share some thoughts.
Any ideas and comments are welcome. Please feel free to discuss about this inequality.
By C-S $$\sum_{cyc}\frac{bc}{\sqrt{3a+1}}\leq\sqrt{\sum_{cyc}bc\sum_{cyc}\frac{bc}{3a+1}}$$ and it's enough to prove that $$\sum_{cyc}bc\sum_{cyc}\frac{bc}{3a+1}\leq\frac{9}{4},$$ which is a linear inequality of $w^3$.
Can you end it now?