Why is Mathematical Induction used to prove solvable inequalities?

As a first year undergrad student I've seen problems where solvable inequalities need to be proven to hold in a specific domain using Mathematical Induction.

My question is, if the inequalities are solvable, i.e. where the bounds where the inequality holds true, it's domain where it holds, can be solved for, why is there a need for induction to prove the inequality holds true?

For example this was one of the problems where I had to prove the inequality held true throughout the domain.

Prove using Mathematical Induction: $$n^2 \geq 2n +3, \forall n\in \mathbb{Z} \geq3$$

But I could far easier solve the inequality for the domain where the inequality is true. As shown below. $$Let, P(n) = n^2 \geq 2n+3$$ $$Let, D = \{n \in \mathbb{Z} | n \geq 3 \}$$ Required to prove: $$P(n) \forall n \in D$$ Proof : $$n^2 -2n -3 \geq 0$$ $$(n-3)(n+1) \geq 0$$ $$(n \geq 3\wedge n \geq -1) \vee (n \leq 3 \wedge n \leq -1)$$ $$(n \geq 3) \vee (n \leq -1)$$ $$n \in \mathbb{R} (-\infty, 1] \cup [3, \infty)$$

Since $$\mathbb{Z} \subset \mathbb{R}$$ we have proven P(n) for all elements in D. $$Q.E.D$$

First off is what I've done above Mathematically, correct, and furthermore is it formally correct? I've written what I set out to prove in symbolic form, which is why some of the proof may seem a bit redundant.

Secondly why would I need to use Mathematical Induction for cases like these where the domains in which the inequality holds true is easily solvable, what is the upside of using Mathematical Induction in this case?

Thirdly I was told that Mathematical Induction is great to use for natural numbers and integers, but that it fails for real numbers. If that is correct wouldn't that favor using the direct solving of the inequality root as I did above, as solving the inequality would solve for the domain in which it is true for up to real and complex numbers if I'm not mistaken.

Finally if you have spotted any errors in my reasoning or proof writing techniques/skills, please feel free to criticize and inform me of it, I'm majoring in Math, so don't hold back on me.

• This example is remarquably easy, because any highschooler can solve a quadratic equality or inequality. Can you solve over $\mathbb{R}$ an inequality involving a polynomial of degree 17 ? – Captain Lama Mar 29 '16 at 17:33
• I think that this sort of questions (prove by math. ind. that...) are mainly asked to students to increase their familiarity with mathematical induction. – drhab Mar 29 '16 at 17:33
• It happens fairly often that an "induction exercise" can be solved more simply without induction. The excuse for assigning such exercises is that induction is important, but one quickly runs out of problems which do not require a long introduction, and for which the induction step is technically straightforward. – André Nicolas Mar 29 '16 at 17:35
• @Captain Lama I understand that this is a ridiculously easy example, and I do realize that Mathematical Induction does become useful for proving inequalities that are not directly solvable (such as this one), but for inequalities that are easily solvable, such as this one, is there any benefit to using Mathematical Induction, I wouldn't think so? – Perturbative Mar 29 '16 at 17:37
• Non, that's true, for such easy inequalities, the main interest is scholastic ; you have to start easy with your students. :) – Captain Lama Mar 29 '16 at 17:40