Induction step: $5 + 5n \leq {n}^2$ for $n \geq 6$

Question

Prove by mathematical induction that $5 + 5n \leq {n}^2 $ for all integers $n\geq 6$.

Step 1: Base case

Suppose $n = 6$, hence $5 + 5(6) \leq {6}^2 = 35 \leq 36$

We proved that base case is true as 35 is less than or equal to 36.

Step 2: Induction step

We claim that k is true for some integer more than or equal to 6, therefore $5 + 5k \leq {k}^2$ (*)

We now need to prove that k+1 claim is true and that is $5 + 5(k + 1) \leq {(k + 1)}^2$

I am stuck at this step. Somehow I am unable to sub in my claim k which is the (*) equation into my k+1 equation correctly. Is the step up to this point correct?

I have tried expanding out the k+1 equation for the LHS to get $5k + 10$ but it looks absolutely wrong. If I do it to the RHS, I get $7k + 6$ which looks wrong too although I am able to sub in ${k}^2$ to the RHS equation.

Can someone please tell me how to proceed from here on?

Answer 1

\begin{align} 5+5(k+1) &= 5+5k + 5 \\ &\le k^2+5 \\ &\le k^2 + 2k+1 \\ &=(k+1)^2 \end{align}

The second last step is due to $5 \le 2k+1$ which is equivalent to $2 \le k$, we know this is true since $k \ge 6$.

Answer 2

For the induction step, you assume $5+5k\le k^2$.

Now add $2k+1$ to both sides: $$5+5k+(2k+1)\le (k+1)^2.$$ But $k\ge 6$, so certainly $2k+1\ge 5$.

Hence $5+5(k+1)\le (k+1)^2$.

Answer 3

Alternatively, rewrite $5 + 5n \leq n^2$ as $f(n)=n^2- 5n -5 \ge 0$. Then $ f(n+1)-f(n)=2(n-2) \ge 0 $.