Proof by induction that $3^n \geq 2n^2 + 3n$ for $n \ge 4$ Problem:
If $n$ is a natural number and $n\geq4$, then $3^n \geq 2n^2 + 3n$. (Prove by Induction.)
Attempt at solution:
1) Given: $n$ is a natural number, $n \geq 4$.
2) Let $P(n)$ be the statement "$3^n \geq 2n^2 + 3n$."
3) $P(4) = 3^4 > 2(4)^2 + 3(4)$, i.e. $81 > 44$, thus the base case of $P(4)$ is true.
4) Assume $P(k) = 3^k \geq 2k^2 + 3k$ is true, where $k$ is a natural number and $k \geq 4$.
5) $P(k + 1) = 3^k(3) = 3^{k+1} > (2k^2 + 3k)(3) = 6k^2 + 9k$.  
And this is where I am stuck. I know that I am trying to get $P(k+1)$ into the form $3^{k+1} > 2(k+1)^2 + 3(k + 1)$, and I have tried all sorts of algebraic manipulation, but I still am nowhere close to arriving at the correct form. How should I proceed?
 A: Try expanding $2(k+1)^2 + 3(k+1)$ in order to compare it with $6k^2 + 9k$ (I.e., what you need to ensure that $6k^2 + 9k \geq 2(k+1)^2 + 3(k+1))$:
$$\begin{align} 2(k+1)^2 + 3(k+1) & = 2(k^2 + 2k + 1) + 3k + 3 \\ &= 2k^2 + 4k + 2 + 3k + 3 \\ &= 2k^2 + 7k +5\end{align}$$
Now, all that remains (while keeping in mind that $\bf k\geq 4)$ is to show that $$\begin{align} 6k^2 + 9k & \geq 6k^2 + 7k + 5 \\ 
& \geq 2k^2 + 7k + 5 \\ 
& = 2(k+1)^2 + 3(k+1)
\end{align}$$
So indeed, with the base case, and the fact that $P(k) \implies P(k+1)$, we have established that $P(n)$ holds for all $n \geq 4$.
A: Remember that $k \geq 4$. Then:
$$6k^2 + 9k \geq 2k^2 +9k \geq 2k^2 +7k + 5 = 2(k+1)^2 + 3(k+1)$$
A: From $3^{n}\geq2n^{2}+3n$ it must be proved that $3^{n+1}\geq2\left(n+1\right)^{2}+3\left(n+1\right)$. 
(i.e. $P(n)\Rightarrow P(n+1)$)
Equivalently from $3^{n+1}\geq3\left(2n^{2}+3n\right)$ it must be
proved that $3^{n+1}\geq2\left(n+1\right)^{2}+3\left(n+1\right)$.
That comes to proving that $3\left(2n^{2}+3n\right)\geq2\left(n+1\right)^{2}+3\left(n+1\right)$. 
This for $n\geq4$.
Can you do that?
A: For the induction step we have , assuming the induction hypotheses : $3^k \ge 2k^2+3k$, that :

$3^k(3)=3^{k+1}>(2k^2+3^k)(3)=6k^2+9k$

which, due to the fact that $k \ge 4$, is :

$\ge 6k^2+7k+8 \ge 2k^2+7k+5 = 2(k+1)^2+3(k+1)$.

A: $3^n\geq2n^2+3n$
(1) $n=4$ 
$81\geq44$
(2) Assume true for $n=k$ ,Therfore we have :
$3^k\geq2k^2+3k$
(3) Then for $n=k+1$
$3^{k+1}\geq2(k+1)^2+3k+3$
Simplifying this gives: 
$3(3^k)\geq2k^2+7k+5$
Now to prove this ^^^ you can subsititute what we assumed in (2) to show this inequality stands.
$3^k\geq2k^2+3k$
So..
$3(2k^2+3k)\geq2k^2+7k+5$ 
$6k^2+9k\geq2k^2+7k+5$ 
$6k^2+7k+2k\geq2k^2+7k+5$ 
^^^ This is true, therefore its true for n=k+1 
So there you have it
