<s>[INST] Show that for all positive real numbers $a,b$, not both of $a(1-b),b(1-a)$ are greater than $\frac{1}{4}$ Question: 

Show that for all positive real numbers $a,b$, not both of $a(1-b),b(1-a)$ are greater than $\frac{1}{4}$

Attempt:
I have attempted several things with this problem. I will note what I feel is the most promising: note that $a(1-b)=\frac{1}{4}$ and $b(1-a)=\frac{1}{4}$ induces the following functions: 
$$y=1-\frac{1}{4x} \ \ \ and \ \ y=\frac{1}{4(1-x)}$$
As a side note, these functions are inverses. I noticed that these functions intersect at $x=\frac{1}{2}$ and if we had chosen any value other than $\frac{1}{4}$ we would've ended up with more than one  or no intersections. So I thought let $c$ be a real positive number and suppose  $a(1-b),b(1-a)$ are both greater than $c$. Then the induced functions are: $$y=1-\frac{c}{x} \ \ \ and \ \ y=\frac{c}{1-x}.$$We have to show that the set $$\{x>0,y>0,y>1-\frac{c}{x},y>\frac{c}{1-x}\}$$ is empty when $c>\frac{1}{4}$. We note that the functions intersect at $$x=\frac{1}{2}(1\pm\sqrt{1-4c})$$ which is clearly not real when $c>\frac{1}{4}.$
 [/INST] Suppose for some positive real numbers $a,b$ we have $a(1-b) > \dfrac{1}{4}$ and $b(1-a) > \dfrac{1}{4}$.
Then, $a(1-b) \cdot b(1-a) > \dfrac{1}{4} \cdot \dfrac{1}{4}$, which is equivalant to $a(1-a) \cdot b(1-b) > \dfrac{1}{4} \cdot \dfrac{1}{4}$. 
Is this possible?
</s>