How prove this $a_{n}>1$ let $0<t<1$, and $a_{1}=1+t$, and such
$$a_{n}=t+\dfrac{1}{a_{n-1}}$$
show that $a_{n}>1$
My try: since 
$$a_{1}=1+t>1$$
$$a_{2}=t+\dfrac{1}{a_{1}}=t+1+\dfrac{1}{1+t}-1>2\sqrt{(t+1)\cdot\dfrac{1}{1+t}}-1=2-1=1$$
$$a_{3}=t+\dfrac{1}{a_{2}}=t+\dfrac{1}{t+\dfrac{1}{t+1}}=t+\dfrac{t+1}{t^2+t+1}=1+\dfrac{t^3+t}{t^2+t+1}>1$$
$$\cdots\cdots\cdots$$
But $a_{n}$ is very ugly,so this problem may use other methods.Thank you very much!

In fact,we have 
$$1<a_{n}<1+t$$
we can use  Mathematical induction to prove it
