Prove that $n^2 > n+1 \quad\forall n \geq 2$ using mathematical induction

Question

Prove $n^2 > n+1$ for $ n \geq 2$ using mathematical induction

So I attempted to prove this, but I'm not sure if this is a valid proof.

Base case, $n = 2$ $$ 2^2 > 2+ 1 $$ $n = k + 1$, induction on $n$ $$ (k+1)^2 > k + 2 $$ $k^2 + 2k + 1 > k + 2$ and by assumption, we know that $k^2 > k + 1$. so we can now write: \begin{gather} k^2 + 2k + 1 > 2k + 2\\ k^2 + 1 > 2\\ \end{gather} $k^2 -1 > 0$ and since $n$ has to be $\geq 2$, the left hand side of the inequality will always be positive, making the statement a tautology.

Does this prove the original statement? If not, how do you do it?

Answer 1

We have $$\begin{align} k^2&\ge k+1\tag{1}\\ k^2+2k&\ge 3k+1\tag{2}\\ k^2+2k+1&\ge 3k+2\tag{3}\\ k^2+2k+1&\ge 3k+2\ge k+2\tag{4}\\ (k+1)^2&\ge (k+1)+1\tag{5}\\ \end{align}$$

Answer 2

Suppose it is true for $n=k(\ge 2)$ i.e. $k^2>k+1$

For $n=k+1$

$(k+1)^2=k^2+1+2k>(k+1)+1+2k=3k+2>k+2$

Answer 3

Use the fact that $2^2 > 2+1$ and then that for $n \geq 1 , d/dx(x^2)=2x> d/dx(x)=1$

Answer 4

facts:

• Since $n>2$, $n^2>4>3$
• Induction hyp: $n^2>n+1$

Proof: $(n+1)^2=n^2+2n+1>3+2n+1=2n+4=2(n+2)>n+2=(n+1)+1$.

Answer 5

consider

$\begin{align}(k+1)^2&=k^2+1+2k\\ &>k+1+1+2k\\ &=k+2+2k\\ &>k+2\quad (\because k\geq 2) \end{align}$

Answer 6

$k^2>k+1\\ \implies k(k-1)>1\\ \implies k((k+1)-2)>1\\\implies k(k+1)>2k+1>1\\\implies(k+1)((k+1)-1)>1\\ \implies (k+1)^2>(k+1)+1$