Prove this inequality by induction

Question

I am having trouble proving the following by induction, in particular implementing the IH

$$2^n > 3n^2 +3n +1~\forall n\in \mathbb{N},n\geq8$$

Answer 1

First prove $$3n^2>3n+6 \hspace{0.5cm} \text{for} \thinspace n \geq 3 \hspace{1cm} (1),$$ by induction .

Suppose $$ 3n^2>3n+6,$$ then $$ 3(n+1)^2=3n^2+6n+3>3n+6+6n+3=9n+9>3n+9=3(n+1)+6.$$

Now we prove $2^n > 3n^2+3n+1$ for all $n \geq 8$. Base case $n=8$ is just evaluating.

Now, induction hypothesis: $$ 2^n > 3n^2+3n+1.$$ Then \begin{equation*} \begin{aligned} 2^{n+1}=2 \cdot 2^n >2(3n^2+3n+1)&=6n^2+6n+2 \\ &=3n^2+3n^2+6n+2 \\ &> 3n^2 +3n+6+6n+2 \hspace{0.5cm} \text{by} \thinspace (1)\\ &= 3n^2+9n+8 \\ &= 3(n+1)^2+3(n+1)+2. \end{aligned} \end{equation*}

Answer 2

We prove by induction the following statement is true: $$2^n > 3n^2 +3n +1~\forall n\in \mathbb{N},n\geq8 \tag{1}$$

We shall re-arrange the statement to fit the following, equivalent form:

$$\frac{2^n}{n^2} > 3\left(1+\frac{1}{n}+\frac{1}{3n^2}\right)\tag{2}$$

When $n = 8$, you can verify the statement is true.

Assume the re-arranged, equivalent, and without loss of generality the same equation is true for $n \in \Bbb{N}$. Is it true for $n+1$?

Substituting yields $\frac{2\cdot2^n}{(n+1)^2} > 3 + \frac{3}{n+1} + \frac{1}{(n+1)^2}$.

Also equivalent:

$$\frac{2^n}{n^2} > 3\left(\frac{1}{2} + \frac{9}{2n} + \frac{5}{2n^2}\right) \tag{3}$$

Subtracting $(2)$ from $(3)$ yields the following fact: $(2)$'s RHS is larger than $(3)$'s RHS for $n \geq 8$, and the difference keeps increasing, i.e. $(3)$'s RHS will always be smaller than $(2)$'s RHS, for $n \geq 8$. Therefore, the statement is true for $n+1$.

Answer 3

Since $2^n$ grows faster than $3n^2 + 3n + 1$, this is really straightforward.

in particular implementing the IH

It looks like this:

$$\begin{array} {rll} 2^n &> 2n^2 + 3n + 1 \text{ and } n \ge 8 & \text{Inductive Hypothesis} \\ &~~\vdots \\ 2^{n + 1}&> 2(n + 1)^2 + 3(n+1) + 1 & \text{To Establish} \end{array}$$

Note that your inductive hypothesis is actually 2 assumptions, $2^n > 2n^2 + 3n + 1$ and $n \ge 8$. You have to fill in the $\dots$ with an argument that assumes the hypothesis and leads to the conclusion. A first simplification would be to multiply both sides of the assumption by 2 :

$$\begin{array} {rll} 2^n &> 2n^2 + 3n + 1 \text{ and } n \ge 8 & \text{Inductive Hypothesis} \\ 2^{n + 1} &> 4n^2 + 6n + 2 & \text{Multiply both sides by 2} \\ & ~~ \vdots \\ 2^{n + 1} &> 2(n + 1)^2 + 3(n+1) + 1 & \text{To Establish} \end{array}$$

Now you assume that $A > B$, and you want to prove $A > C$. So attempt to establish that $B > C$ (you'll have to use $n \ge 8$).

$$\begin{array} {rll} n & \ge 8 & \text{Given} \\ & ~~ \vdots \\ 4n^2 + 6n + 2 & > 2(n + 1)^2 + 3(n+1) + 1 & \text{To Establish} \\ \end{array}$$

Work backwards, simplifying:

$$\begin{array} {rll} n & \ge 8 & \text{Given} \\ & ~~ \vdots \\ 2n^2 - n - 4 &> 0 & \text{To Establish} \\ 4n^2 + 6n + 2 &> 2n^2 + 4n + 2 + 3n+3 + 1 & \text{Algebra} \\ 4n^2 + 6n + 2 &> 2(n + 1)^2 + 3(n+1) + 1 & \text{Algebra} \\ \end{array}$$

Since the coefficient of $n^2$ (the fastest growing term) is positive (2), after a point it will always be positive. One thing you could do is find the minimum of the quadratic, which occurs at $4n - 1 = 0$ or $n = 1/4$. Since the polynomial is increasing after $n=1/4$, and is positive at $n = 8$, you can conclude $2n^2 - n - 4 > 0$ which is the last unestablished proposition.

Alternatively, you could establish $\forall n \ge 8 ~:~ 2n^2 - n - 4 > 0$ with induction (I won't show the base case):

$$\begin{array} {rll} 2n^2 - n - 4 &> 0 \text{ and } n \ge 8 & \text{Inductive Hypothesis} \\ & ~~ \vdots \\ 2(n + 1)^2 - (n + 1) - 4 & > 0 & \text{To establish} \\ \end{array}$$

Same as before, use algebra etc to finish the proof.

Answer 4

Here's a general hint. Suppose we are working to show that $p_d(n) < 2^n$ for all $n \geq N$, where $p_d$ is a degree $d$ polynomial (you have a quadratic in your case).

For our inductive step we assume $p_d(n) < 2^n$ and we then want to show that $p_d(n+1) < 2^{n+1}$ follows.

We can write this as $p_d(n+1) = p_{d}(n) + p_{

This problem is very similar to the one you started with, and once we solve it we've solved the original problem.
We'll do this by induction, checking the base case $n = N$ and then working through the inductive step. So what's happening is we are getting a recursive chain of inductions, which must terminate because each time we dive into the next step, our polynomial has strictly lower degree than before. Eventually we must get to the problem $C = p_0(n) < 2^n$, which is verifiable immediately (there's no induction to be had).
For your case of comparing a quadratic polynomial to $2^n$, your induction step will give you a linear function to compare with $2^n$, and finally a constant which you'll see is less than $2^8$.

The cool thing about looking at the process abstractly is that it quickly leads to a proof (by induction) that all polynomials exhibit rate of growth lower than exponential.