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reenmachine

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## 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!

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