1. Proof Question

There exists a c in R+ st. for all n in N, n>=2 --> 3n^2 - 4n >= cn^2
How would I approach a proof idea, then find an appropriate value for c and then a formal well structured proof. I intuitively figured out c <= 2 && c > 1, But how to begin?

Thanks
Laz

2. Originally Posted by Lazarath
There exists a c in R+ st. for all n in N, n>=2 --> 3n^2 - 4n >= cn^2
How would I approach a proof idea, then find an appropriate value for c and then a formal well structured proof. I intuitively figured out c <= 2 && c > 1, But how to begin?

Thanks
Laz
You need to find a $\displaystyle c\in \mathbb{R}^+$ such as, $\displaystyle 3n^2-4n\geq cn^2$ for all $\displaystyle n\geq 2$. Notice that,
$\displaystyle 3n^2-4n\geq cn^2$
if and only if, (divide by $\displaystyle n\geq 2$) thus, postive the sign does not change
$\displaystyle 3n-4\geq cn$
if and only if,
$\displaystyle 3n-cn\geq 4$
if and only if,
$\displaystyle n(3-c)\geq 4$
if and only if, for all $\displaystyle n\geq 2 \in \mathbb{Z}$
$\displaystyle 3-c\geq \frac{4}{n}$ (1)

If $\displaystyle a<b$ and (1) is true for $\displaystyle a$ then it is true for $\displaystyle b$ too. Thus, if (1) is true for the smallest $\displaystyle n$ then it is true for all integers bigger than it. Thus, it must be true for 2.
Thus, $\displaystyle 3-c\geq \frac{4}{2}=2$
Thus, solving for $\displaystyle c\leq 1$.

Now we check whether $\displaystyle c\leq 1$ is a solution. Thus, $\displaystyle 3-c\leq 2=\frac{4}{2}$. Since it is true for $\displaystyle n=2$ it is true for all $\displaystyle n\geq 2$ as explained in the previous paragraph.

$\displaystyle \mathbb{Q}.\mathbb{E}.\mathbb{D}.$

3. Thank you

Just wanted to say thanks for the timely reply. Never thought I'd be in University of Toronto at 40yrs trying to wrap my head around this!

Thanks Again,

Lazarath