The sequence I am working with is:

xn+1=xn(x2n+15)3x2n+5x_{n+1} = \frac {x_n(x_n^2+15)}{3x_n^2+5}

Here we know that x1>0x_1>0

I know that this sequence converges to √5\sqrt{5}. I would prove it inductively if I was given an initial value, knowing that if x1>√5x_1>\sqrt{5}, then the sequence would be bounded below by √5\sqrt{5}, and if x1<√5x_1<\sqrt{5}, then the sequence would be bounded above by √5\sqrt{5}. The problem is that I do not know how to show that this converges, without the ability to prove it inductively.
Any help here would be greatly appreciated. Thanks in advance.
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Try to prove that|x2n+1−5|≤K|x2n−5||x_{n+1}^2-5|\leq K |x_n^2-5| with 0