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).$

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 $a-\frac 12$.
