Determine the minimum possible value of the sum \(\frac{a}{2b} + \frac{b}{4c} + \frac{c}{8a},\)

where \(a,b\) and \(c\) are positive real numbers.

Havingfun Aug 4, 2019

#1**+1 **

\(\text{There are a couple ways to do this. We'll do it formally first, by setting the gradient to 0}\\ f=\dfrac{a}{2b}+\dfrac{b}{4c}+\dfrac{c}{8a}\\ \nabla f = \left(\dfrac{1}{2b}-\dfrac{c}{8a^2},~\dfrac{1}{4c}-\dfrac{a}{2b^2},~\dfrac{1}{8a}-\dfrac{b}{4c^2}\right)\\ \nabla f=0 \Rightarrow\\ \dfrac{1}{2b}=\dfrac{c}{8a^2}\\ \dfrac{1}{4c}=\dfrac{a}{2b^2}\\ \dfrac{1}{8a}=\dfrac{b}{4c^2}\)

\(\text{The only solution with no negative values is $b=2a,~c=2a$}\\ \text{This results in $f = \dfrac 3 4$ but more importantly it results in $~\\$ each of the terms contributing equally to the sum}\\ f = \dfrac{a}{4a}+\dfrac{2a}{8a} + \dfrac{2a}{8a} = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4\)

\(\text{You can use that symmetry principle to come up with the answer in a simpler way}\\ \text{Just find the relationship between the variables that makes each term of what you're trying to $~\\$ find the extrema of contribute the same amount}\\ \text{Here you'd say that $\dfrac{a}{2b} = \dfrac{b}{4c} = \dfrac{c}{8a}$}\\ \text{and solve for $b$ and $c$ in terms of $a$ as we did above}\)

.Rom Aug 4, 2019