I am trying to prove something then I hit a stumbling block in the form of this problem.
Let $a,b,c,d$ be non-negative integers such that:
$c,d$ are fixed
$\max a=d$
$$a\geq b$$
$$c\geq b$$
$$d\geq a$$
$$d> c$$
I want to show that $a\geq c$ and here is my idea.
I consider the difference $a-c$. This difference is either positive, negative, or zero.
If the difference is positive or zero then I can say that $a\geq c$.
If the difference is negative then that implies that $a<c$ but since $a\leq d$ this will imply that $d$ can be less than $c$ which is a contradiction since $d>c$.
So I can only have that $a-c$ is positive or zero thus $a\geq c$.
Is my reasoning correct here? Can you suggest an alternate or correct proof? Thank you!
Your reasoning is false.
Try $a=2, b=1, c=3, d=4$. Then $d \gt c \gt a \gt b$ and this satisfies your four original inequalities but $a \not \ge c$.
Your error is in saying "If the difference is negative then that implies that $a<c$ but since $a\leq d$ this will imply that $d$ can be less than $c$ which is a contradiction since $d\geq c$" as it only leads to a contradiction if $d$ must be less than $c$, and that is not the case.