Q:

Binomial theorem with inequality

I was reading the resolution of a problem and came across the following:
$\dfrac{T_{p+1}}{T_p}=\dfrac{121 - p}{2p}; T_{p+1} > T_p$ if $p \leq 40$ and $T_{p+1} < T_p$ if $p \geq 41$
for $p \in \mathbb{N}$ and $p \geq 0$.
Can anyone help me clarify these inequalities?
The problem is:
Determine the maximum term in the development of $(1 + \dfrac{1}{2})^{120}$

A:

Notice that $f(p):=\frac{121-p}{2p}=1$ for $p=40+\frac{1}{3}$. You can show that $0<f(p)<1$ for $120\ge p\ge 41$ and $f(p)>1$ for $0<p\le 40$. Therefore, 

If $0< p\le 40$, $1< \frac{T_{p+1}}{T_p}$. 
If $41\le p\le 120$, $0< \frac{T_{p+1}}{T_p}<1$. 

If you multiply the two inequalities by $T_p$ you will get the desired result but you have to prove that $T_p>0$ for $0<p\le 120$ which can be done by induction. 

