Determine whether there are rational numbers $a, b, c$ such that $3=a+b+c=\frac{1}{(a^2-1)}+\frac{1}{(b^2-1)}+\frac{1}{(c^2-1)}$

Question

I have tried using $AM\ge HM$ inequality for ${(a^2-1)},{(b^2-1)},{(c^2-1)}$ which gives me $$ \frac{(a^2-1)+(b^2-1)+(c^2-1)}{3} \ge \frac{3}{\frac{1}{(a^2-1)}+\frac{1}{(b^2-1)}+\frac{1}{(c^2-1)}}$$ $$ \frac{a^2+b^2+c^2-2(a+b+c)+3}{3} \ge \frac{3}{\frac{1}{(a^2-1)}+\frac{1}{(b^2-1)}+\frac{1}{(c^2-1)}}$$

Since $ a+b+c=\frac{1}{(a^2-1)}+\frac{1}{(b^2-1)}+\frac{1}{(c^2-1)}=3$, $$ \frac{a^2+b^2+c^2-6+3}{3} \ge 1$$ $$ {a^2+b^2+c^2} \ge 6$$

However, I can't move on from that.