max of min of max of min Can I somehow prove or disprove that $R' \le R$. What assumptions, if any do I need for this?
$R$ and $R'$ defined as follows.
\begin{align}
R & = \max \biggl(0, \min\Bigl(R_{full},K_1 + \max\bigl(0, \min (K_2, R_{full})\bigr)\Bigr)\biggr) \\
R' & = \max \biggl(0, \min\Bigl(R_{full},K_1 + \delta + \max\bigl(0, \min (K_2-\delta, R_{full})\bigr)\Bigr)\biggr)
\end{align}
Here, 
$R_{full}$ and $\delta$ are strictly positive values.
The only difference between $R$ and $R'$ is that of $\delta$.
My initial impulse was to try to make a table of all possible choices ($2$ per $\min$ or $\max$ so $2^4$ in total), and for everyone of those, show that same result holds (ie $R' \le R$ or vice versa). But this does not seem right. 
Can any one point me to a better way?
Thanks.
 A: Note that $R' \nleq R$. To see that, let's presume $K_2 = R_{full}+\delta$, $K_1 = -\delta$, and $0 < \delta < R_{full}$. Then
$$R = \max(0,\min(R_{full},R_{full}-\delta))=R_{full}-\delta,$$
and
$$R' = \max(0,\min(R_{full},R_{full}))=R_{full}.$$
Clearly, $R' > R$.
A: For $a=\max (0,x)$ and $b=\max (0,y)$, if $y\geq x$ then $b\geq a$. Therefore, one has to show if $$\min\Bigl(R_{full},K_1 + \delta + \max\bigl(0, \min (K_2-\delta, R_{full})\bigr)\Bigr) \geq \min\Bigl(R_{full},K_1 + \max\bigl(0, \min (K_2, R_{full})\bigr)\Bigr)$$
holds true.
For $x=\min (a,b)$ and $y=\min (a,c)$, if $b\geq c$ then $x\geq y$
Therefore one must check if $$K_1 + \delta + \max\bigl(0, \min (K_2-\delta, R_{full})\bigr)\geq K_1 + \max\bigl(0, \min (K_2, R_{full})\bigr)$$
or 
$$\delta + \max\bigl(0, \min (K_2-\delta, R_{full})\bigr)\geq  \max\bigl(0, \min (K_2, R_{full})\bigr)\tag 1$$
holds true. Consider the case $K_2-\delta<R_{full}$ and $K_2>R_{full}$. In this case the right side of the inequality will be equal to $R_{full}$ because, $R_{full}$ is given as strictly positive. The left side can take the values from the set $\{\delta,K_2\}$. 
Lets assume $\max(0,K_2-\delta)=0$, this means $K_2-\delta<0$ and $K_2<\delta$. Now for this case the left side of the equation $(1)$ is $\delta$, we know that $\delta>K_2$ and our initial condition was that $K_2>R_{full}$. Hence, $\delta>R_{full}$ and the inequality holds.
Lets assume $\max(0,K_2-\delta)=K_2-\delta$. Then, the left side of the inequality will be $K_2$. When we look back to our initial assumption that $K_2>R_{full}$, we see that $(1)$ also holds in this case.
For every $K_2<R_{full}$ we have $K_2-\delta<R_{full}$ and for all these cases we have $\min (K_2-\delta, R_{full})=K_2-\delta$ and $\min (K_2, R_{full})=K_2$.
Therefore, either we have $\max(0,x)=\max(0,y)$ where $x=y=R_{full}$ for which $(1)$ holds with $>$ or we have $\max(0,x)=K_2-\delta$ and $\max(0,y)=K_2$
In this case $(1)$ simplifies to 
$$\delta + \max\bigl(0, K_2-\delta\bigr)\geq  \max\bigl(0, K_2\bigr)\tag 2$$
We know that $K_2>K_2-\delta$ hence, if $K_2<0$ so is $K_2-\delta$ and in this case $(2)$ holds with $>$.
if $K_2-\delta<0$ and $K_2>0$ we have $$\delta\geq^? K_2$$ For this case we asssumed that $K_2-\delta<0$ therefore $K_2<\delta$ and the inequality holds again.
The last case is that both $K_2-\delta>0$ and $K_2>0$, In this case $(1)$ holds with equality.
Hence, we conclude that $R^{'}\geq R$ for all cases.
