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

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!!

• After you got $$x_0 +\frac{n^2}{x_0-x_n}\ge x_n+2n$$ its just AM-GM on $x_0-x_n$ and $\dfrac{n^2}{x_0-x_n}$. Anyways, isn't this problem in your blog? Jan 2, 2021 at 10:20
• @Anand yeah but that's AM-GM only :( idk why this pr is supposed to use Titu , which I think I haven't used nicely . if u get one do post the sol Jan 2, 2021 at 12:26
• We are using $T_2$ as well...Like $$(x_0-x_n)+\left(\sum_{i=0}^{n-1}\frac{1}{x_{i}-x_{i+1}}\right)\overset{AM-GM}{\geq}2\sqrt{(x_0-x_n)\underbrace{\left(\sum_{i=0}^{n-1}\frac{1}{x_{i}-x_{i+1}}\right)}_{\overset{T_2~lemma}{\geq}\frac{n^2}{x_0-x_n}}}\geq 2\sqrt{n^2}=2n$$ Jan 2, 2021 at 14:07

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

• oh wow! I got it !!! thankyou :) it follows because $a+1/a\ge 2$ . Jan 2, 2021 at 4:21
• @SunainaPati You are welcome! Jan 2, 2021 at 4:22

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.

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

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