POTW #291: Minimum Value of a Tricky Expression | MathHelpBoards

[anemone]

Here is this week's POTW:

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

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[anemone]

Congratulations to castor28 for his correct solution(Smile), which follows:

Spoiler

Let us write $S$ for the expression in question, and $x = a-b$, $y=b-c$; this implies that $a-c=x+y$. We have:

$$
\begin{align*}\displaystyle
S &= \frac{1}{x}+\frac{1}{y} + \frac{1}{x+y}\\
&= \frac{x+y}{xy} + \frac{1}{x+y}
\end{align*}
$$
​

Because $xy(x+y)=17$, we may write this as:

$$
\begin{align*}\displaystyle
S &= \frac{(x+y)^2}{17} + \frac{1}{x+y}\\
&= \frac{z^2}{17} + \frac{1}{z}
\end{align*}
$$
​

with $z=x+y>0$.

The derivative is:

$$\displaystyle
S' = \frac{2z}{17} - \frac{1}{z^2}
$$
​

which has a single positive root at $z=\sqrt[3]{17/2}\approx 2.0408$; the value of S at that point is $\dfrac{3}{2z}\approx 0.7350$.

This is indeed a minimum, since $S\to +\infty$ for $z\to0$ and $z\to +\infty$.