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.
Thank you in advance.
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.
Thank you in advance.
Inductive Hypothesis:
$\displaystyle \left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)^k>\frac{k}{2}$
Show:
$\displaystyle \left(\frac{5}{4}\right)^{k+1}-\left(\frac{3}{4}\right)^{k+1}>\frac{k+1}{2}$
Working with the left side...
$\displaystyle \left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^ k-\left(\frac{5}{4}\right)\left(\frac{3}{4}\right)^k$
$\displaystyle \left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\frac{5}{4}\left[\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)^k\right]$
On the right hand side we have 5/4 times our inductive hypothesis, substituting we get...
$\displaystyle \left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\frac{5}{4}\left(\frac{k}{2}\right)=\left(1+\frac {1}{4}\right)\left(\frac{k}{2}\right)=\frac{k}{2}+ \frac{k}{8}>\frac{k}{2}+\frac{1}{2}$
$\displaystyle \frac{k}{8}>\frac{1}{2}$ Because k>4
So, by transitivity, we get...
$\displaystyle \left(\frac{5}{4}\right)\left(\frac{5}{4}\right)^k-\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)^k >\frac{k}{2}+\frac{1}{2}$
$\displaystyle W^5$ "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.
Given that:
$\displaystyle \left(\frac{5}{4}\right)-\left(\frac{3}{4}\right)^x>\left(\frac{x}{2}\right )$
Show:
$\displaystyle 3^x+4^x<5^x$
Using the given inequality... (multiply both sides by $\displaystyle 4^x$, note that this is legal because exponential functions are non-negative)
$\displaystyle 4^x\left[\left(\frac{5}{4}\right)-\left(\frac{3}{4}\right)^x>\left(\frac{x}{2}\right )\right]$
$\displaystyle 5^x-3^x>4^x\frac{x}{2}$
Adding $\displaystyle 3^x$ to both sides we get...
$\displaystyle 5^x>4^x\frac{x}{2}+3^x>4^x+3^x$
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
$\displaystyle 5^x>4^x+3^x$
Once again, hope that made sense...