Determine the minimum possible value of the $\dfrac{a}{2b} + \dfrac{b}{4c} + \dfrac{c}{8a}$,where $a,b,$ and $c$ are positive real numbers.
How should I go about writing equations to solve for values that can satisfy this expression? Is there another way to do this besides writing system of equations?
Apply AM-GM inequality to the sum: $S \ge 3\sqrt[3]{\dfrac{a}{2b}\cdot \dfrac{b}{4c}\cdot \dfrac{c}{8a}}= \dfrac{3}{4}$ which is the minimum.This value of $\dfrac{3}{4}$ occurs when $\dfrac{a}{2b} = \dfrac{b}{4c} = \dfrac{c}{8a}\implies b^2 = 2ac, c^2 = 2ab\implies b^3 = c^3 = 2abc\implies b = c= 2a.$
Of course the AM-GM inequality is the smart way to solve the problem : https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means
But, if you are not comfortable with this method you can find the extrema (minima and/or maxima) thanks to differential calculus. $$f(a,b,c)=\frac{a}{2b} + \frac{b}{4c} + \frac{c}{8a}\tag 1$$ $$\frac{\partial f}{\partial a}=\frac{1}{2b} - \frac{c}{8a^2}=0 \tag 2$$ $$\frac{\partial f}{\partial b}=-\frac{a}{2b^2} + \frac{1}{4c}=0 \tag 3$$ $$\frac{\partial f}{\partial c}=-\frac{b}{4c^2} + \frac{1}{8a}=0 \tag 4$$ From $(2)$ we get $c=\frac{4a^2}{b}$. Putting it into $(3)$ and $(4)$ leads to $8a^3-b^3=0$. The result is : $$b=2a=c$$ $$f(a,2a,2a)=\frac{a}{2(2a)} + \frac{(2a)}{4(2a)} + \frac{(2a)}{8a}=\frac34$$ This shows that $f$ has only one finite extremum$=\frac34$. Is it a maximum or a minimum ?
Compute any other value of $f\:$, for example $f(1,1,1)=\frac12+\frac14+\frac18=\frac78$
The function $f(a,b,c)$ is continuous for $a,b,c>0$ and $\frac78>\frac34$ thus $\frac34$ is the minimum.
