and leads to r>1/2.
It works for \(\large a=1~\&~r=\frac{1}{6}\)I was trying to prove using a>0 but not the given result. The above can be demonstrated to be >0 if a>0 after simplifying, and leads to r>1/2.
The only thing \((2r-1)^2>0\) says is that for every \(r \ne\frac{1}{2}\) that is a statement.
Can you multiply both sides by \(\large a~?\)I'm sorry I can't.
Use \( r=\dfrac{1}{2}\) in the expression \(\dfrac{a}{1-r}>4ar\).So r cannot equal 1/2 is the exact value asked for?