Question

The sum of the squares of three distinct real numbers, which are in $G.P$. is $S^2$. If their sum is $\alpha S$, show that ${\alpha }^2\in (\frac{1}{3},1)\cup (1,3)$

Answer 1

Let the numbers be $ar,a,a/r$ such that
$a(r+1+\frac{1}{r})=\alpha S$
and $a^2(r^2+1+\frac{1}{r^2})=S^2$
Put $r+\frac{1}{r}=t$ $\therefore r^2+\frac{1}{r^2}=t^2-2$
$\therefore a(t+1)=\alpha S$ and $a^2(t^2-1)=S^2$
Eliminating $S$, we get $a^2(t^2-1)=\frac{a^2{(t+1)}^2}{{\alpha }^2}$
$\therefore (t-1){\alpha }^2=(t+1)$ or $t=\frac{{\alpha }^2+1}{{\alpha }^2-1}$
Now $t=r+\frac{1}{r}$ $\therefore r^2-rt+1=0$
For $t$ be real $t^2-4>0$ $\therefore (t+2)(t-2)>0$
$\therefore t<-2,t>2$
Hence from $\left(1\right)$, we get
$\frac{{\alpha }^2+1}{{\alpha }^2-1}<-2$ or $\frac{{\alpha }^2+1}{{\alpha }^2-1}>2$
or $\frac{{\alpha }^2+1}{{\alpha }^2-1}+2<0$ or $\frac{{\alpha }^2+1}{{\alpha }^2-1}-2>0$
Note : In an inequality we can multiply only by a $+ive$ quantity. Here we do not know whether ${(\alpha }^2-1)$ is $+ive$ or $-ive$
or $\frac{3{\alpha }^2-1}{{\alpha }^2-1}<0$ or $\frac{3{-\alpha }^2}{{\alpha }^2-1}>0$ $\Rightarrow \frac{{\alpha }^2-3}{{\alpha }^2-1}<0$
$\therefore \frac{3({\alpha }^2-\frac{1}{3})({\alpha }^2-1)}{{({\alpha }^2-1)}^2}<0$
and $\frac{({\alpha }^2-1)({\alpha }^2-3)}{{\left({\alpha }^2-1\right)}^2}<0$
The second implies that ${\alpha }^2$ lies between $1/3$ and first implies that ${\alpha }^2$ lies between $1/3$ and $1$
$\therefore {\alpha }^2\in (\frac{1}{3},1)\cup (1,3)$.