Basic Algebra Proof on Integers - Weak Inequalities Work but Strict Inequalities Don't? 
Let $a, b, \& \, m$ be integers. Prove that if $2a + 3b \geq 12m + 1$, then $a \geq 3m + 1$ or $b \geq 2m + 1$.

My Attempt: I don't conceive apace how to contrive, from the one inequality in the antecedent, the two inequalities in the consequent. So a proof by contraposition may be more facile. The contrapositive is: $\text{If }  \color{Green}{a < 3m + 1} \; \& \; \color{#0073CF}{b < 2m + 1}, \text{ then } 2a + 3b < 12m + 1 \tag{*}.$
From $(*)$, $\color{Green}{2a} + \color{#0073CF}{3b} < \color{Green}{6m + 2} \; + \;\color{#0073CF}{6m + 3} = 12m +5$. But this doesn't prove $(*)$.

Given Solution:  Assume that $a < 3m+1$ and $b < 2m+1.$ Since a and b are integers, $a ≤ 3m$ and $b ≤ 2m.$ Therefore, $2a + 3b ≤ 2(3m) + 3(2m) = 12m < 12m+ 1,$ as desired. $\blacksquare$

What went wrong in my attempt? I understand the given solution but I persisted with the more natural and direct green and blue, instead of deviating to weak inequalities. Shouldn't both solutions work?
Source: Problem 4.19 on P102 (related to P90, Result 4.8) of Mathematical Proofs, 2nd ed by Chartrand et al.

$\large{\text{Supplement to Cameron Buie and pritam's Answers :}}$
What overriding, catholic lessons can be extrapolated from this example to generalise about inequalities? 
Moreover, in order to circumvent the problem here (ie the loss of information from the weaker inequalities), should I always start with the strictest inequality? If so, should I always rework any weak inequality into the equivalent strict inequality? 
For example, define $r \in \mathbb{R} $ and $n \in \mathbb{Z}$ and $g(...) \neq h(...)$. Then $ f(r) < h(r) \require{enclose}
     \enclose{updiagonalstrike}{\iff} f(r) \leq g(r).$
But $ f(n) < h(n) \text{ MAY OR MAY NOT }\iff f(n) \leq g(n).$
 A: What went wrong with your solution was that you didn't use the fact that $a$, $b$ and $m$ are integers. If they were real numbers the inequality $2a+3b\ge12m+1$ would not imply the conclusion given.
For example, if $m=1$, $a=3.5$, $b= 2.5$ we have $2\times 3.5 + 3\times 2.5 = 7+7.5=14.5 \ge 13$.
The solution given does use the fact that the numbers are integers - the next integer down from $a$ is $a-1$, and you can't have $a-\frac 12$.
A: I will try to explain why your attempt doesn't work. From your assumptions $a<3m+1$ & $b<2m+1$ you can write $$a=3m+1-x\quad\&\quad b=2m+1-y$$ where $x,y>0$. Then , $$2a+3b=12m+5-(2x+3y)$$ Now note that, if you don't use the fact that $x,y$ are integers, then you only get $2x+3y>0$, which is what you did. If you use the fact that $x,y$ are integers then you will get $$2x+3y\geq2\cdot 1+3\cdot 1=5$$ so $2a+3b\leq 12m+5-5=12m$.
A: As requested, I will expound upon my comment, but I'll do it in the form of an answer, since it's rather lengthy.
I noted that the given proof deviates to stronger inequalities, which allow the proof to work. By that I mean, for example, that if $a$ and $m$ are integers, then $a<3m+1$ if and only if $a\le 3m.$ They are not equivalent among real numbers (take $m=1$ and $a=3.5,$ for example), but the inequality $a\le 3m$ implies the inequality $a<3m+1$, meaning $a\le 3m$ is the stronger of the two inequalities, despite being non-strict.
What if we multiply both inequalities by $2$? Then we're dealing with $2a\le 6m$ and $2a<6m+2.$ These are still equivalent if we know that $a,m$ are integers, but it isn't as obvious why, since one of them is misleading. The kicker is divisibility. We already know that there are no integers in the interval $(6m+1,6m+2)$ since $m$ is an integer, which lets us rewrite $2a<6m+2$ equivalently as $2a\le 6m+1$. Since $a$ and $m$ are integers, though, we have that $2a$ is even and $6m+1=2(3m)+1$ is odd, so we can't have equality. Hence, we rewrite it equivalently as $2a<6m+1,$ which in turn yields $2m\le 6m.$
Similarly, taking $b,m$ to be integers, $3b<6m+3$ yields $3b\le 6m+2;$ since $3$ divides $3b$ and does not divide $6m+2=3(2m)+2,$ then we have $3b<6m+2,$ which in turn yields $3b\le 6m+1;$ but $3$ does not divide $6m+1=3(2m)+1,$ so this yields $3b<6m+1,$ which in turn yields $3b\le 6m.$
Why did the strengthened forms of the inequalities work when the others didn't? Well, quite simply because the "weaker" versions weren't precise enough--they were suggesting more room to maneuver than there actually was, so once we "added the inequalities together" we had an inequality that was too weak, and didn't clearly imply our desired result. Indeed, as you saw, once we put $2a<6m+2$ and $3b<6m+3$ together to get $2a+3b<12m+5,$ we're out of luck. That extra pad in the strict inequalities left us with a $5$ that we can't get rid of--simple divisibility arguments won't let us remove it the way we did above, because it isn't clear what numbers should divide $2a+3b$. However, with the "strictest" versions of the inequalities, we simply get $2a+3b\le12m,$ which readily yields the desired conclusion.
