Is this how you prove by induction for inequalities? the question is here: http://cpsc.ualr.edu/srini/DM/chapters/examples/ex2.3.2.html
My solution is as below:

 A: Your method is valid. One thing to keep in mind is that though Induction solves the problem, it does so only for integers. A much simpler proof, would be $$ n^2 \ge 3n \ge 2n + 3 $$
Also, make sure you keep a distinction between what you want to prove, and what you already know. An better layout would be as such:
The base case: $3^2 \ge 2(3) + 3$
The inductive step: Suppose $k^2 \ge 2k + 3$. Adding $2k + 1$ to both sides gives $$k^2+ 2k + 1 \ge 4k+4 $$ We know $4k + 4 \ge 2k + 10 \ge 2(k+1) + 3$, so $$ (k+1)^2 \ge 4k + 4 \ge 2(k+1) + 3 $$which completes our inductive step.
A: Prove true for $n = k + 1$.
$${(k + 1)}^2 \ge 2(k + 1) + 3.......a$$
Taking the $LHS$ of inequality $a$ you end up with
$${(k + 1)}^2 = k^2 + 2k + 1......(1)$$
Using the assumption, $k^2 \ge 2k +3$, $(1)$ becomes, $${(k + 1)}^2 = k^2 + 2k + 1 \ge (2k + 3) + 2k + 1 (\forall n \ge 3)$$
$${(k + 1)}^2 \ge (2k + 3) + 2k + 1.........b$$
Taking the RHS of inequality $b$,
$$(2k + 3) + 2k + 1 = 4k + 4 = 4k + 1 + 3 = 2 \cdot 2k + 1 + 3$$
$$2 \cdot 2k + 1 + 3 \ge 2(k + 1) + 3 (\forall n \ge 3)$$
Therefore $${(k + 1)}^2 \ge 2(k + 1) + 3 (\forall n \ge 3)$$
