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!

Let $\displaystyle \mu = \frac{t + \sqrt{t^2+4}}{2}$, we have
$$\mu > 1\quad\text{ and }\quad\mu(t - \mu) = \left(\frac{t + \sqrt{t^2+4}}{2}\right)\left(\frac{t - \sqrt{t^2+4}}{2}\right) = -1$$
From this, we get
$$a_{n+1} -  \mu = t - \mu + \frac{1}{a_n} = \frac{1}{a_n} - \frac{1}{\mu} = \frac{\mu - \alpha_n}{\mu a_n}$$
This implies if $a_n > 1$, then 
$$|a_{n+1}-\mu| = \frac{|a_n - \mu|}{\mu a_n} < \frac{|a_n -\mu|}{\mu} < |a_n - \mu|\tag{*1}$$
Notice
$$\begin{align}
( 1 - \mu)^2 - (a_1 - \mu)^2 = & (1 - \mu)^2 - (1 + t - \mu)^2 = (1 - \mu)^2 - (1 - \frac{1}{\mu})^2\\
= & (1-\mu)^2(1 - \frac{1}{\mu^2}) > 0
\end{align}$$
We have $a_1 \in (1,2\mu - 1) = (\mu - (\mu - 1),\mu + (\mu - 1))$. Since all $x$ in this interval $> 1$, we can repeatedly apply $(*1)$ to conclude all $a_n$ belongs to this interval and hence $> 1$.
