More specifically, consider two real numbers $a,b>0$, and their average $r=\frac{a+b}{2}$.
It is the case that $a=r*x$ and $b=r*y$ where $\vert 1-x\vert =\vert 1-y\vert$.
For example, let $a=5$ and $b=7$. Then $r=6$, and $a= .833333 r$ and $b=1.1666666 r$.
How can I formally prove this?
More specifically, is there a way to prove this that isn't by doing the cases $a>b$ and $a<b$ separately?
I ask about doing it without cases because we can write $a= \frac{2a}{a+b} \frac{a+b}{2}$ and $b= \frac{2b}{a+b} \frac{a+b}{2}$
so then we just need to show the relationship between $ \frac{2a}{a+b}$ and $ \frac{2b}{a+b}$
if $a>b$ then $ \frac{2a}{a+b} -1 = \frac{a-b}{a+b} = 1- \frac{2b}{a+b}$
but if $a<b$ then we need to write $ 1- \frac{2a}{a+b} = \frac{b-a}{a+b} = \frac{2b}{a+b}-1$
i.e. depending on if $a>b$ or $b<a$ we write $ \frac{2a}{a+b} -1$ or $ 1- \frac{2a}{a+b} $. Is there a way I can do this proof without cases?
You can use the fact that $$|X|=|Y|\iff X=\pm Y$$ So, we have $$\begin{align}&|1-x|=|1-y| \\\\&\iff 1-x=1-y\quad\text{or}\quad 1-x=-(1-y) \\\\&\iff x=y\quad\text{or}\quad x+y=2\end{align}$$