MHB What is the Minimum Value of a Mathematical Function with Specific Constraints?

[anemone]

Find the minimum of $\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{a-c}$ for real $a>b>c$ given $(a-b)(b-c)(a-c)=17$.

 

[Opalg]

Find the minimum of $\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{a-c}$ for real $a>b>c$ given $(a-b)(b-c)(a-c)=17$.

[sp]Since the problem only depends on the differences between the numbers, we may as well add $-c$ to each of them, so that $c$ becomes $0$.

Then we want to minimise $S \overset{\text{def}}{=} \dfrac1{a-b} + \dfrac1a + \dfrac1b$ subject to the constraint $ab(a-b) = 17.$

Write the constraint as $ab^2 - a^2b + 17 = 0$ and consider it as a quadratic in $b$. Its discriminant is $a^4 - 68a$, and this must be non-negative if there is to be a real solution for $b$. Therefore $a^3 - 68 \geqslant0$, or $a\geqslant \sqrt[3]{68}.$

Next, $S = \dfrac{ab + (a+b)(a-b)}{ab(a+b)} = \dfrac{a^2 + ab-b^2}{17} = \dfrac{a^2}{17} + \dfrac{a^2b - ab^2}{17a} = \dfrac{a^2}{17} + \dfrac1a.$ For $a>0$, this has its minimum value when $a = \sqrt[3]{17/2}.$ But that is less than $\sqrt[3]{68}.$ So the minimum value of $S$ occurs when $a=\sqrt[3]{68},$ and $S$ is then equal to $\dfrac5{\sqrt[3]{68}}.$[/sp]