Prove that $x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}\ge x_n+2n$

Question

St.Petersburg 1999 :Let $x_0>x_1>\dots >x_n$ be real numbers. Prove that $$ x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}\ge x_n+2n $$

I tried Titu's inequality.

So we get $$ x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}= x_0+\frac{1^2}{x_0-x_1}+\frac{1^2}{x_1-x_2}+\dots+\frac{1^2}{x_{n-1}-x_n} \ge x_0 +\frac{n^2}{x_0-x_n}.$$

So we need to show that $$x_0 +\frac{n^2}{x_0-x_n}\ge x_n+2n .$$

Now note that $$\frac{n^2}{x_0-x_n}=\frac{n^2}{x_0}+\frac{x_n \cdot n^2}{x_0\cdot(x_0- x_n)}$$

So we need to show that $$x_0 +\frac{n^2}{x_0}+\frac{x_n \cdot n^2}{x_0\cdot(x_0- x_n)}\ge x_n+2n .$$

Note that by AM-GM $$x_0 +\frac{n^2}{x_0}\ge 2n$$.

So it's enough to show that $\frac{x_n \cdot n^2}{x_0\cdot(x_0- x_n)}\ge x_n$ .

Enough to show that $$\frac{n^2}{x_0\cdot(x_0- x_n)}\ge 1.$$

Which I am not able to. Any hints? Thanks in advance!!

Answer 1

Hint: let $x_0=x_1+a_1,x_1=x_2+a_2....x_{n-1}=x_n+a_n$ where $a_i>0$

we have to prove $$x_{0}-x_{n}+\frac{1}{a_1}+\frac{1}{a_2}...+\frac{1}{a_n}\ge 2n$$ $$a_1+a_2+a_3+.....a_n+\frac{1}{a_1}+\frac{1}{a_2}..+\frac{1}{a_n}\ge 2n $$ $$\iff \left(a_1+\frac{1}{a_1} \right)+\left(a_2+\frac{1}{a_2} \right)......+\left(a_n+\frac{1}{a_n} \right)\ge 2n$$ which is true by AM-GM

Answer 2

Note that your last inequality need not be true, esp for $ x_0 > n$, so unfortunately that's futile.
The reason being that in the previous inequality, equality holds iff $ x_0 =n$. And when $ x_0 > n$, we have some leeway to work with, that is needed for your last inequality.

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Instead, do you see that you can prove this, similar to what Albus stated?

So we need to show that $$x_0 +\frac{n^2}{x_0-x_n}\ge x_n+2n .$$