We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

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.

 Aug 4, 2019

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


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


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

 Aug 4, 2019

21 Online Users