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Let $x_0>x_1>x_2>\ldots>x_n$ be real numbers.

Prove $\displaystyle x_{0}+\frac{1}{x_{0}-x_{1}}+\frac{1}{x_{1}-x_{2}}+\ldots+\frac{1}{x_{n-1}-x_{n}}\geq x_{n}+2n$.

I thought I would try and deal with the denominators by letting $a_k=x_{k-1}-x_k$

Which gives:

LHS $\displaystyle=x_0+\frac{1}{a_1}+\frac{1}{a_2}+\ldots +\frac{1}{a_n}$

I am unsure how to use the AM-GM mean from here.

Anthony Macks
  • 229

2 Answers2

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Required to prove:

$\displaystyle x_{0}+\frac{1}{x_{0}-x_{1}}+\frac{1}{x_{1}-x_{2}}+\ldots+\frac{1}{x_{n-1}-x_{n}}\geq x_{n}+2n$

i.e., To prove that (TPT) $$\sum a_k + \sum \frac{1}{a_k} \geq 2n$$ where $a_k=x_{k-1}-x_k$ for $k = 1, \ldots, n$

i.e., TPT $$nA + \frac{n}{H} \geq 2n$$ where $A = \frac{\sum a_k}{n}$ is the arithmetic mean and $H = \frac{n}{\sum \frac{1}{a_k}}$ is the harmonic mean

i.e., TPT $$A + \frac{1}{H} \geq 2$$

However, $$A + \frac{1}{H} \geq A + \frac{1}{A} \geq 2$$

Hence the proof.

PTDS
  • 3,464
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Thanks PTDS. I saw your solution and had a light bulb moment. Recall that the AM-GM mean for two(2) numbers can give us: $\displaystyle a+\frac{1}{a}\ge2$

Restate the original problem as:

{$\displaystyle x_0-x_n+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\ldots +\frac{1}{x_{n-1}-x_n}\ge2n$}

Now let $a_k=x_{k-1}-x_k$

So, $x_0 - x_n = a_1 + a_2 + \ldots + a_n$

Thus,

LHS $=\displaystyle a_1+a_2+\ldots+a_n+\frac{1}{a_1}+\frac{1}{a_2}+\ldots +\frac{1}{a_n}$

$\displaystyle =a_1+\frac{1}{a_1}+a_2+\frac{1}{a_2}+\ldots a_n+\frac{1}{a_n}$

$\ge 2+2+\ldots +2$ (n times)

$=2n$

$=$RHS

Anthony Macks
  • 229
  • Nice solution ! using the fact that a+1/a \ge 2 was awesome ! – Sunaina Pati Dec 19 '20 at 05:09
  • A side-note: if we do not stick to the AM-GM line, the AM-HM gives us a stronger (with "larger" lower bound) version of the inequality: $LHS \ge \frac{n^2}{x_0-x_n} + {x_0-x_n} \ge RHS$ – Czylabson Asa Aug 04 '21 at 14:54