A geometric series modified from AMC2016

$\begin{aligned} a_3 &= a_0 \cdot q^2 = 1 \\ \implies q &\neq 0 \\ \therefore S &= \dfrac {a_0}{1-q} \\ &= \dfrac {a_0 \cdot q^2}{(1-q)q^2} \\ &= \dfrac {a_3}{4(1-q)(\frac q2)(\frac q2)} \\ &\geq \dfrac 1{4\left (\dfrac {1-q+\frac q2 + \frac q2}{3} \right)^3} \\ &= \dfrac {27}{4} \end{aligned}$

Let the first term be $a$ and the common ratio be $R$ . Then:

$S = \frac{a}{1-R}$ $T_3 = 1 = aR^2$ $S = \frac{T_3}{R^2(1-R)}$ $\implies S = \frac{1}{R^2(1-R)}$ For $S$ to be minimum, the denominator should be as high as possible. In other words:

$f = R^2 - R^3$

Should be maximised.

$\implies \frac{df}{dR} = 2R-3R^2=0$ $\implies R = 0 \ \mathrm{or} \ R = \frac{2}{3}$

The solution $R=0$ corresponds to a local minimum as the second derivative at that point is positive. $R=\frac{2}{3}$ corresponds to a local maximum. The value of $S$ corresponding to this is:

$S =\frac{27}{4}$