Proving the Inequality: 0<r<1 Implies 1/(r(1-r))≥4 | Help with Basic Proof

[dustbin]

Homework Statement

Prove that if r is a real number such that 0<r<1, then

$$\frac{1}{r(1-r)} \geq 4.$$

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,

$$\frac{1}{r(1-r)} = \frac{1}{\frac{1}{n}(1-\frac{1}{n})} = \frac{n^2}{n-1} \geq 4.$$

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.

$$\frac{2^2}{2-1} = 4 \geq 4$$

Now we let n=k such that $$\frac{k^2}{k-1} \geq 4$$ and show that the inequality is true for k+1. Observe that

$$\frac{(k+1)^2}{(K+1)-1} = \frac{k^2+2k+1}{k} \geq 4$$ which we can rearrange as $$k^2 \geq 4k-2k-1 = 2k-1 = 2(k-\frac{1}{2}) .$$

From the induction hypothesis we see that $$k^2 \geq 4k-4 = 4(k-1)$$ which shows that the inequality is true for k+1. Thus $$\frac{n^2}{n-1} = \frac{1}{r(1-r)} \geq 4$$ 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!

 

[Villyer]

It can be easier to manipulate the original inequality into a quadratic equation, and then find the minimum value on the range (0,1) to show that it is always larger than 4.

 

[who_]

As Villyer said, simply manipulating the inequality from what you are given will do the trick. Also, just because r is a real number DOES NOT mean that r = 1 / n for some integer n. That is not even true for rational numbers, so the proof you have right now is technically incorrect.

[EDIT] Actually, now that I have read your "proof" more carefully, it is very incorrect since you use induction, which relies on the assumption that r = 1/n for some integer n, which is false.

 

[dustbin]

Ah. I again over-complicate my search! So now I have:

Let rεR such that 0<r<1. Then we have
$$\frac{1}{1(1-r)} \geq 4 <br /> \longleftrightarrow 1 \geq 4r-4r^2<br /> \longleftrightarrow 0 ≥ -4r^2+4r-1<br /> \longleftrightarrow 0 ≥ -(2r-1)^2 <br /> \longleftrightarrow 0 ≤ (2r-1)^2.$$

/ end formal argument!

I know that r= 1/2 here and that this is the minimum of the range between 0<r<1, which can be shown with first/second derivative tests. What is necessary for me state in order to show that 4 is the minimum value of the range on 0<r<1? Is there a simpler way for me to verify that this is the minimum value?

@who_:
I see my flaw! A number 0<r<1 could not only be 1/n but also 3/9, sqrt(6)/99... I don't know where I got that idea from

 

[who_]

That's a much nicer and cleaner proof :)

Haha, no worries - even the most talented mathematicians make mistakes like those.