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Given $f(x)=\sqrt{x^2+1}, g(x)=\sqrt{\frac{3x^2-1}{x^4-1}}$, prove that $f(x)=g(x)$ for $x
eq \pm \sqrt{1/3}$

I want to prove that given $f(x)=\sqrt{x^2+1}, g(x)=\sqrt{\frac{3x^2-1}{x^4-1}}$, $f(x)=g(x)$ for $x
eq \pm \sqrt{1/3}$.
I’ve got a pretty good idea, but I’m not exactly sure how to prove it. I’ll outline my thoughts.
Suppose not, then there is some non-positive real number $c$ (for $x
eq \pm \sqrt{1/3}$) such that $f(x)>g(x)$ for all $x \in [0, c)$, but $f(x)0$ then this implies that $f$ and $g$ are not equal for any $x \in (-\infty, c)$. Clearly $f(0)=g(0)=0$, hence $f$ and $g$ are not equal for $x=0$, it also follows that $f(x)$ and $g(x)$ are not equal for any other $x$.
I was thinking of doing some bounds for $f$ and $g$ to work out where their graphs intersect, but I haven’t seen a bound that has been proven on how many times the graphs intersect for $x \in (0, +\infty)$.
Is there some sort of interval I could prove such that I only need to consider the endpoints of

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