Prove by mathematical induction

Question

I stuck with a problem like this. I know all the steps but I can't prove that it is true when n=k+1.

n^2 ≥ 2n + 1, for all n ∈ N such that n ≥ 3.

Answer 1

Direct algebric proof: write the inequality as $x^2-2x-1 \geq 0$. The roots are $x=1 \pm \sqrt{2}$, so that the inequality is true for $x \leq 1-\sqrt{2}$ or $x \geq 1+\sqrt{2}$. Since $x$ is assumed to be a natural number, we have that the inequality holds for $x \geq \lceil 1+ \sqrt{2} \rceil$, and then $x\geq 3$.

Induction proof: the inequality is clearly true for $x=3$. Now we have to show that, if the inequality holds for some integer $x$ (i.e., $x^2\geq 2x-1$), it also holds for $x+1$, which means $(x+1)^2\geq 2(x+1)-1$.

To do this, let us rewrite the last inequality as

$$x^2+2x+1 \geq 2x+2+1$$

Since we are assuming that $x^2 \geq 2x-1$, the inequality above remains true if we rewrite the LHS as $(2x+1)+(2x+1)=4x+2$. This yields

$$4x+2 \geq 2x+2+1$$

and then

$$2x \geq 1$$

which is clearly true for all natural numbers.