Simplifying a complication max operation I have dervied an inequality and have arrived to the following 
$$\max\{1, \frac{b}{2}+1\} \leq \max\{a, \frac{b}{2}+ \frac{a}{2}\}$$
I am trying to simplify further and arrived to the following conclusions 
$$a \geq 1$$
$$ a \geq \frac{b}{2}+1$$
$$ a\geq 2 $$
$$\frac{b}{2}+ \frac{a}{2} \geq 2 $$
How can I proceed and further simplify, are these inequalities redundant?
Thanks
 A: Here is a graph of your inequality, with $a$ the horizontal axis and $b$ the vertical axis.

We can see the strange shape of the boundary line: it is $a=1$ for $b\le 0$ and $a=2$ for $b\ge 2$, with the line segment from $(1,0)$ to $(2,2)$. There are several ways to describe this more simply than your maxima, but one way is
$$a\ge 2 \text{ or } (a\ge 1 \text{ and } b\le 2a-2)$$
Another way is
$$\begin{cases}
a\ge 1,  & \text{if $b\le 0$} \\[2 ex]
a\ge \frac 12b+1,  & \text{if $0<b<2$} \\[2 ex]
a\ge 2,  & \text{if $b\ge 2$} \\
\end{cases}$$
You can see that your conclusions are not quite correct. For example, it is not always true that $a\ge 2$, since $a=1,b=0$ satisfies your inequality.
A: Starting with
$$\max\{2, \frac{b}{2}+1\} 
\leq \max\{a, \frac{b}{2}+ \frac{a}{2}\}
$$
Let's look at the left side.
Case 1:
$2 > \frac{b}{2}+1
$.
Then
$b < 2$.
This then becomes
$2 
\le \max\{a, \frac{b}{2}+ \frac{a}{2}\}
< \max\{a, \frac{1}{2}+ \frac{a}{2}\}
= \max\{a,  \frac{a+1}{2}\}
$.
If
$a > \frac{a+1}{2}
$,
then
$a < 1
$
so that
this becomes
$2 < 1$
which is false.
Therefore,
$a \le \frac{a+1}{2}
$,
so that
$a \ge 1
$
so that
this becomes
$2 < \frac{a+1}{2}$
or
$a > 1
$
which we already know.

Case 2:
$2 \le \frac{b}{2}+1
$.
Then
$b \ge 2$.
This then becomes
$\frac{b}{2}+1 
\le \max\{a, \frac{b}{2}+ \frac{a}{2}\}
=\frac{a}{2}+ \max\{\frac{a}{2}, \frac{b}{2}\}
$.
If
$a \ge b
$,
this becomes
$\frac{b}{2}+1 
\le a
$.
But since
$b \ge 2$,
$b \ge \frac{b}{2}+1 $
which is true.
If $a < b$,
this becomes
$\frac{b}{2}+1
< \frac{a}{2}+\frac{b}{2}
$
or
$a > 2$.
In all cases here,
$a \ge 2$.

Therefore,
the solutions are
$b < 2, a > 1$;
$b \ge 2, a \ge 2$.
