- #1
dustbin
- 240
- 5
Homework Statement
Prove that if r is a real number such that 0<r<1, then
[tex] \frac{1}{r(1-r)} \geq 4.[/tex]
The Attempt at a Solution
Assume that r is a real number such that 0<r<1. Then r= 1/n where nεZ such that n≥2. As such,
[tex] \frac{1}{r(1-r)} = \frac{1}{\frac{1}{n}(1-\frac{1}{n})} = \frac{n^2}{n-1} \geq 4. [/tex]
It can be shown by induction that this inequality is true for n≥2 where n is an integer.
First the base case is proven where n=2.
[tex] \frac{2^2}{2-1} = 4 \geq 4 [/tex]
Now we let n=k such that [tex] \frac{k^2}{k-1} \geq 4 [/tex] and show that the inequality is true for k+1. Observe that
[tex] \frac{(k+1)^2}{(K+1)-1} = \frac{k^2+2k+1}{k} \geq 4 [/tex] which we can rearrange as [tex] k^2 \geq 4k-2k-1 = 2k-1 = 2(k-\frac{1}{2}) .[/tex]
From the induction hypothesis we see that [tex] k^2 \geq 4k-4 = 4(k-1) [/tex] which shows that the inequality is true for k+1. Thus [tex] \frac{n^2}{n-1} = \frac{1}{r(1-r)} \geq 4 [/tex] for nεZ such that n≥2.
I apologize if this is a bit hard to read... this my first attempt at latex! I realize that this only takes into account rational numbers and disregards the rest of the reals. I made this attempt hoping I would see a way to complete the proof, but I am drawing a blank on how to include the rest of the real numbers. Any suggestions are greatly appreciated!