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# Help

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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

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

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

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

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Aug 4, 2019