Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24} + \frac{12}{17} = \frac{577}{408}$$
The sequence seems decreasing to me so I tried to prove it by induction. Need to prove $p_k - p_{k+1} \gt 0$ for all n.
For n = 1, $p_1 - p_2 = 2- \frac{3}{2} = \frac{1}{2} \gt 0$ (true)
However when I tried to prove it for $k+1$, I ran into problems.
Assume $p_k - p_{k+1} \gt 0$ is  true for n =k, then it must be also true for $n =k+1$.
$$p_{k+1} - p_{k+2} = \frac{p_k}{2} + \frac{1}{p_k} - \frac{p_{k+1}}{2} - \frac{1}{p_{k+1}} = \frac{p_k - p_{k+1}}{2} + (\frac{1}{p_k}-\frac{1}{p_{k+1}})$$
but $\frac{1}{p_k}-\frac{1}{p_{k+1}} \lt 0$
I don't know what to do from there.

AM/GM. $\frac{x}{2}+\frac{1}{x}<x$ iff $x^2>2$. So it is enough to show that if $x^2>2$ then $(\frac{x}{2}+\frac{1}{x})^2>2$ or $\frac{x^2}{4}+\frac{1}{x^2}>1$. But by AM/GM $(\frac{x^2}{4}+\frac{1}{x^2})/2>\sqrt{\frac{1}{4}}$.
