# Proof exercise

1. Jul 31, 2013

### reenmachine

1. The problem statement, all variables and given/known data

Prove that if $x \in ℝ$ and $0<x<4$ , then $\frac{4}{x(4-x)}≥1$.

(I'm really not sure about that one but managed to attempt something)

2. Relevant equations

We first want to prove that $\forall y \in \mathbb{Z}$ , the largest product of any $a,b \in ℝ$ with $a+b=y$ is when $a=b=\frac{y}{2}$.

Suppose $a=b=\frac{y}{2}$.We have $\frac{y}{2} \cdot \frac{y}{2} = \frac{y^2}{4}$.

Now suppose $a≠b≠\frac{y}{2}$.This implies that $\exists x \in ℝ \ \ a=(\frac{y}{2}-x) \ \ b=(\frac{y}{2}+x)$ with $a$ and $b$ being interchangeable.We have $(\frac{y}{2}-x)(\frac{y}{2}+x) = \frac{y^2}{4}-x^2$.Since $\frac{y^2}{4} > \frac{y^2}{4}-x^2$ , this proves that $\forall y \in \mathbb{Z}$ , the largest product of any $a,b \in ℝ$ with $a+b=y$ is when $a=b=\frac{y}{2}$.

To get back at our original proof , we know that $x+(4-x)=4$.From our proof above , we know that the largest product from any $a,b \in ℝ$ such that $a+b=4$ is if $a=b=\frac{4}{2}$.We have $2 \cdot 2 = 4$.This gives us $\frac{4}{4}=1$.Since all other products of any other $a,b \in ℝ$ will result in a smaller number , we know that $\frac{4}{4-n} \ \ n \in \mathbb{R} = 1+c$ with $c \in R^+$.This proves that if $x \in ℝ$ and $0<x<4$ , then $\frac{4}{x(4-x)}≥1$.

any help greatly appreciated!!! thank you!

Last edited: Jul 31, 2013
2. Jul 31, 2013

### lurflurf

3. Jul 31, 2013

### reenmachine

Unfortunately I don't know about these concepts yet.I just tried to solve it on top of my head.Thank you for answering! Does this mean that the proof is incorrect , or is it simply a clearer way of proving it?

4. Jul 31, 2013

### lurflurf

Your proof is correct (except I am slightly confused by your y why is it important that it is an integer and is it defined by a+b=y or ab<=y^2/4 (of course both are true, but one is assumed and the other shown)) . As I mentioned you proof is the same as writing x(4-x)=4-(x-2)^2. In that form it is easy to see that x(4-x)=4-(x-2)^2<=4 with equality when x=2. We can also generalize slightly and write
$$ab=\left( \frac{a+b}{2} \right) ^2 - \left( \frac{a-b}{2} \right) ^2 \le \left( \frac{a+b}{2} \right) ^2$$
we see that a=b is required for equality.

5. Jul 31, 2013

### verty

My method for questions like this one is to simplify the formula hoping to reverse the steps later. In this case it works brilliantly.

6. Jul 31, 2013

### Mandelbroth

Can you use calculus? This seems like an easy place to use derivatives to just show the local minimum on $x\in(0,4)$ is greater than or equal to 1.

Your proof seems alright though. There are a few parts where I'm a little hesitant, though. Your notation toward the end implies that $n$ is a real number equal to $1+c$ for some positive real number $c$. Clear these kinds of things up and you should be fine.

7. Jul 31, 2013

### reenmachine

No I didn't learn calculus yet unfortunately.Perhaps this is the wrong section for me to be in , but I figured writing proof didn't belong in the pre-calculus section.

(I edited the n in N for n in R at the end but meant n in R+)