Proving: ##\frac{4}{x(4-x)}≥1## for ##x \in ℝ## and ##0<x<4##

[reenmachine]

Homework Statement

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)

Homework 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!

 

[lurflurf]

http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

$$x(4-x) \le \left( \frac{x+(4-x)}{2}\right) ^2$$

an ad-hoc method in keeping with what you started is to write

$$x(4-x)=4-(x-2)^2$$

 

[reenmachine]

http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

$$x(4-x) \le \left( \frac{x+(4-x)}{2}\right) ^2$$

an ad-hoc method in keeping with what you started is to write

$$x(4-x)=4-(x-2)^2$$

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?

 

[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.

 

[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.

 

[Mandelbroth]

Homework Statement

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)

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.

 

[reenmachine]

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.

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+)

Thanks for the advices!