For $X_{n+1}=X_n+1/X_n$ with $ X_1=1 $, prove that $X_{100}>14$. Given the sequence:
$$X_{n+1}=X_n+1/X_n \qquad X_1=1, $$ prove that  $X_{100}>14$
 A: Observe that $$x_{n+1}^2=x_n^2+2+\frac1{x_n^2}$$ Thus, you might notice that $$x_{n+2}^2=x_{n+1}^2+2+\frac1{x_{n+1}^2}=\left(x_{n}^2+2+\frac1{x_n^2}\right)+2+\frac1{x_{n+1}^2}>x_{n}^2+4$$ In fact, if you keep descending and use similar estimations, you will reach to the following conclusion $$x_{n+2}^2>x_1^2+2(n+1)=2n+3$$ You may use induction to prove this more rigorously. We are now done, since this yields $$x_{100}^2>2\cdot 98+3=199>196=14^2$$
A: As suggested in the comments, by induction you may prove that:
$$ (X_n)^2 \geq 2n-1 $$
Base case $n=1$:
$$1^2 \geq 2-1=1.$$
(IH) Then suppose  for some $k\in \mathbb N_{\geq 1}$ we have that:
$$ (X_k)^2 \geq 2k-1$$
We will now make the inductive step. Observe that by definition of the sequence we have:
$$ (X_{k+1})^2 =(X_k + 1/X_k)^2 =(X_k)^2 + 2 \frac{X_k}{X_k} + \frac{1}{(X_k)^2} \geq (X_k)^2 + 2 .$$
By the induction hypothesis we have:
$$ (X_{k+1})^2 \geq (X_k)^2 + 2 \geq (2k-1)+2 = 2(k+1)-1. $$
Then by the principle of mathematical induction the estimate holds for all $n \in $$\mathbb N_{\geq 1}$, in particular for $n=100$ we have that:
$$(X_{100})^2  \geq 200-1=199 > 196=14^2$$
Hence $X_{100}>14$.
