1. Mathematical Induction Question

I was hoping someone could help me with a proof. I have attached an image.
I am going to prove it using induction (rather than well-ordering), and the base cases where n=1,2 and 3 are straightforward.

2. Inductive Hypothesis:

$\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)^k>\frac{k}{2}$

Show:

$\left(\frac{5}{4}\right)^{k+1}-\left(\frac{3}{4}\right)^{k+1}>\frac{k+1}{2}$

Working with the left side...

$\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^ k-\left(\frac{5}{4}\right)\left(\frac{3}{4}\right)^k$

$\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\frac{5}{4}\left[\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)^k\right]$

On the right hand side we have 5/4 times our inductive hypothesis, substituting we get...

$\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\frac{5}{4}\left(\frac{k}{2}\right)=\left(1+\frac {1}{4}\right)\left(\frac{k}{2}\right)=\frac{k}{2}+ \frac{k}{8}>\frac{k}{2}+\frac{1}{2}$

$\frac{k}{8}>\frac{1}{2}$ Because k>4

So, by transitivity, we get...

$\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\frac{k}{2}+\frac{1}{2}$

$W^5$ "Which Was What We Wanted"

Sorry I didn't know how to do the greater than or equal to symbol.

Hope that was clear enough.

3. Originally Posted by Jen
Sorry I didn't know how to do the greater than or equal to symbol.
just type \ge

type \le for less than or equal to

type \ne for not equal to

4. Oooops, lets try this again.

5. Thank you very much Jen. Now I just need to do c using part b as my solution.

6. Given that:

$\left(\frac{5}{4}\right)-\left(\frac{3}{4}\right)^x>\left(\frac{x}{2}\right )$

Show:

$3^x+4^x<5^x$

Using the given inequality... (multiply both sides by $4^x$, note that this is legal because exponential functions are non-negative)

$4^x\left[\left(\frac{5}{4}\right)-\left(\frac{3}{4}\right)^x>\left(\frac{x}{2}\right )\right]$

$5^x-3^x>4^x\frac{x}{2}$

Adding $3^x$ to both sides we get...

$5^x>4^x\frac{x}{2}+3^x>4^x+3^x$

The inequality on the right is possible because x is always greater than or equal to 2 so x/2 is always greater than or equal to 1.

Once again by transitivity we get

$5^x>4^x+3^x$

Once again, hope that made sense...