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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

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    $\begingroup$ Maybe try using the fact that: $$\max\{x,y\} = \frac{(x+y)+|x-y|}{2}$$ $\endgroup$ – user171308 Aug 25 '15 at 22:38
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Here is a graph of your inequality, with $a$ the horizontal axis and $b$ the vertical axis.

enter image description here

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.

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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

$b < 2, a > 1$;

$b \ge 2, a \ge 2$.

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  • $\begingroup$ Your answer implies that $a=1.4,b=1$ is a solution, but it is not. $\endgroup$ – Rory Daulton Aug 26 '15 at 8:33

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