# Thread: Find max value - (Triangles, geometric progressions, greatest integer(floor)function)

1. ## Find max value - (Triangles, geometric progressions, greatest integer(floor)function)

If for $\displaystyle r > 1$, three successive terms of a Geometric Progression with common ratio $\displaystyle r$ represent sides of a triangle, then find the maximum value of $\displaystyle (\lfloor 2r \rfloor + \lfloor - r \rfloor)$

2. ## Triangle Inequality

Take some triangle with side lengths $\displaystyle a,ar,ar^2$ for some $\displaystyle a>0$ and $\displaystyle r>1$. Without loss of generality, let $\displaystyle a=1$ and find the allowable values of $\displaystyle r$ for a triangle to exist:

(i) $\displaystyle 1+r>r^2$ true for all $\displaystyle r>1$ but $\displaystyle r<\phi$ for $\displaystyle \phi$ the golden ratio $\displaystyle \approx 1.618$
(ii) $\displaystyle 1+r^2>r$ true for all $\displaystyle r>1$
(iii) $\displaystyle r+r^2>1$ true for all $\displaystyle r>1$

Therefore, $\displaystyle r\in(1,\phi)$ to satisfy this condition. It can be shown that:

$\displaystyle \lfloor 2r \rfloor + \lfloor -r \rfloor = \left\{ \begin{array}{rcl} 0 & \mbox{if} & 1<r<1.5 \\1 & \mbox{if} & 1.5\leq r<2 \\2 & \mbox{if} & 2\leq r<2\phi \end{array}\right.$

Therefore the max value of this function on the allowable interval is 2.

QED