Finding the sum of an Infinite GP given $a_1 = 1$ and $4a_2 + 5a_3 $ has least value

Question

I solved a question in a certain way that didn't seem right. I would like to know if there is any other way to do the question or if there is any fault in my logic for the question.
The question says to find the sum of a $\infty$ GP such that $a_1 = 1$ and $4a_2 + 5a_3$ has the least value in GP.
My approach:
I solved by first assuming that since $4a_2 + 5a_3$ has the least value thus we can say, $$5r^2 + 4r \lt 1$$ So, $$ 5r^2 + 5r -r -1 \lt 0$$ Which gives us,
$$(5r-1)(r+1) \lt0$$ So, $$r \in \Bigl(-1,\frac{1}{5}\Bigl)$$ But, $$5r^2 + 4r \lt 1 \iff r \lt 0$$ So $$r \in \Bigl(-1,0\Bigl)$$ Now we know, $$\sum ^{\infty}_{r=1} a_r = \frac{a}{1-r}$$ So, since $$r\lt 0$$ we get $$\sum ^{\infty}_{r=1} a_r = \frac{a}{1+r}\; \forall \;r \in \Bigl(-1,0\Bigl)$$ This implies, since $$a = 1$$ That, $$\sum ^{\infty}_{r=1} a_r = \frac{a}{1+r} \lt 1$$ Only one of the options matches this, and so that is how I arrived at the answer.
Please tell me if there is any flaw in the solution and also how to reach at a precise value.
The answer given is $\frac{5}{7}$ This matches my solution but I still felt like I could do with a second opinion.

Answer 1

$$4a_2+5a_3=4r+5r^2$$ This has the least value when the derivative is $0$. $$\frac d{dr}4r+5r^2=4+10r=0 \\ \implies r = \frac{-2}{5}$$

Hence sum of GP is $$\frac{1}{1-\frac{-2}5}=\frac57$$

Alternatively if you don't know any calculus then note that for a quadratic function $f(x)=ax^2+bx+c$, minima is given by $-b/2a$. The midpoint of the roots of equation $f(x)=0$.