It is true that $1 \leq k \leq n$ then $k^2+n+1 \leq 2k(n+1)$?

Question

Problem:

Suppose $k \in \mathbb{N}$ and $n \in \mathbb{N}$. Suppose $1 \leq k \leq n$.

It is true that $k^2+n+1 \leq 2k(n+1)$?

How can I prove it?

My attempt:

I tried using induction on $n$ but actually I am not sure the statement is true.

Answer 1

We want to prove $k^2 + n + 1 \leq 2k(n+1)$. Moving the $n+1$ to the right, this is the same as $k^2 \leq (n+1)(2k-1)$.

Since $k \geq 1$, we have $2k-1 = k+k-1 \geq 1+k-1 = k$. So $$(n+1)(2k-1) \geq (n+1)k \geq (k+1)k > k^2,$$ as required.