Keep yourself careful...

$\lambda = x , S=s$

$\dfrac{a}{r} , a , ar$

$a(r + 1 + \dfrac{1}{r}) = xs$

$a^2( r^2 +1 + \dfrac{1}{r^2}) = s^2$

$r + \dfrac{1}{r} = y , r^2 + \dfrac{1}{r^2}= y^2 - 2$

$a(y + 1) = xs , a^2(y^2 - 1) = s^2$

$Thus , ~ a^2(y^2 - 1) = \dfrac{a^2(y^2 + 1)^2}{x^2}$

$y = \dfrac{x^2 + 1}{x^2 - 1}$

$r + \dfrac{1}{r} = y , r^2 - yr + 1 = 0$

$as~ratio~is~always~real~thus~discriminant>0$

$y^2 - 4>0 , y< -2 ~or~ y>2$

Thus,

$\dfrac{x^2 + 1}{x^2 - 1} < -2 ~or \dfrac{x^2 + 1}{x^2 - 1}>2$

$\dfrac{x^2 + 1}{x^2 - 1} + 2< 0 ~or~\dfrac{x^2 + 1}{x^2 - 1} - 2>0$

$\dfrac{3x^2 - 1}{x^2 - 1} < 0 ~or~ \dfrac{3 - x^2}{x^2 - 1} >0$

$\boxed{\dfrac{3(x^2 - \dfrac{1}{3})(x^2 - 1)}{(x^2 - 1)^2} <0} ~or \boxed{\dfrac{(x^2 - 1)(x^2 - 3)}{(x^2 - 1)^2} < 0}$

$Thus~ x^2= \lambda^2 \in (\dfrac{1}{3} , 1) \cup (1,3)$