Prove the inequality $\sqrt{\frac{a^2+1}{b+c}}+\sqrt{\frac{b^2+1}{a+c}}+\sqrt{\frac{c^2+1}{a+b}}\ge 3$

Let $a;b;c\in R^+$ such that $ab+bc+ca>0$. Prove that $$\sqrt{\frac{a^2+1}{b+c}}+\sqrt{\frac{b^2+1}{a+c}}+\sqrt{\frac{c^2+1}{a+b}}\ge 3$$

I have seen the similar question is $$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$$ my problem is harder than it

My try: Holder: $$LHS^2\cdot \sum _{cyc}\left(a^2+1\right)^2\left(b+c\right)\ge \left(a^2+b^2+c^2+3\right)^3$$

Then we need to prove: $$\left(a^2+b^2+c^2+3\right)^3\ge 9\sum _{cyc}\left(a^2+1\right)^2\left(b+c\right)$$

After full expanding I am stuck because that inequality is not homogeneous, it hard to solve.

Because by AM-GM and C-S we obtain: $$\sum_{cyc}\sqrt{\frac{a^2+1}{b+c}}\geq3\sqrt[6]{\prod_{cyc}\frac{a^2+1}{b+c}}=3\sqrt[12]{\prod_{cyc}\frac{(a^2+1)(1+b^2)}{(a+b)^2}}\geq3\sqrt[12]{\prod_{cyc}\frac{(a+b)^2}{(a+b)^2}}=3$$