Let $a$, $b$, $m$ be integers. Prove that if $2a + 3b \ge 12m + 1$, then $a \ge 3m + 1$ or $b \ge 2m + 1$.   I need help proving this. I am not sure what to do.   Thank you for all of the edits.   I am still unsure on how to word the problem.    Right now.  Proof - By contrapositive.  Assume that a &lt; 3m + 1 and b &lt; 2m + 1 for the integers a, b, and m. Then,  2a + 3b &lt; 12m + 1 = 2(3m + 1) + 3(2m+1) &lt; 12 m + 1  = 6m + 2 + 6m + 3 &lt; 12m + 1 = 12m + 5 &lt; 12m + 1  Which is false because 5 &lt; 1  So next, I would say...  Assume that a &lt; 3m + 1 and b &lt; 2m + 1 for the integers a, b, and m. Then a &lt; 3m &lt; 3m + 1 and b &lt; 2m &lt; 2m + 1.  Then plug in again to show 12m &lt; 12m + 1  Then what?
Let $a, b, \&amp; \, m$ 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: $\text{If }  \color{Green}{a &lt; 3m + 1} \; \&amp; \; \color{#0073CF}{b &lt; 2m + 1}, \text{ then } 2a + 3b &lt; 12m + 1 \tag{*}.$  From $(*)$, $\color{Green}{2a} + \color{#0073CF}{3b} &lt; \color{Green}{6m + 2} \; + \;\color{#0073CF}{6m + 3} = 12m +5$. But this doesn't prove $(*)$.     Given Solution:  Assume that $a &lt; 3m+1$ and $b &lt; 2m+1.$ Since a and b are integers, $a ≤ 3m$ and $b ≤ 2m.$ Therefore, $2a + 3b ≤ 2(3m) + 3(2m) = 12m &lt; 12m+ 1,$ 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) &lt; h(r) \require{enclose}      \enclose{updiagonalstrike}{\iff} f(r) \leq g(r).$ But $ f(n) &lt; h(n) \text{ MAY OR MAY NOT }\iff f(n) \leq g(n).$
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