Telescopic Inequality!. Right?

Algebra Level 5

Determine the largest real number $a$ such that for all real numbers $x_0$ , $x_1$ , $x_2$ $\cdots$ $x_{2017}$ satisfying $0 = { x }_{ 0 } < { x }_{ 1 } < { x }_{ 2 }< \cdots < { x }_{ 2017 }$ , we have

$\large\ \sum _{ k=1 }^{ 2017 }{ \frac { 1 }{ { x }_{ k }-{ x }_{ k-1 } } } \ge a\left( \sum _{ m=1 }^{ 2017 }{ \frac { m + 1 }{ { x }_{ m } } } \right)$

The answer is 0.444.

1 solution

Jon Haussmann
Aug 5, 2017

I don't have a solution, but I want to make some important observations.

Evidently, this problem is based on problem A8 of the 2016 IMO Short List . Based on the solution provided in the link, we can say that $\frac{4}{9} \le a \le \frac{673}{1514}.$ Since $\frac{4}{9} = 0.444 \dots$ and $\frac{673}{1514} = 0.444518 \dots$ , this is a fairly narrow interval.

The original Short List problem is to determine the best constant that works for all $n$ . Finding the best constant that works for individual values of $n$ , like the problem here, is a different problem. For example, for $n = 2$ , the best constant is $\frac{1}{2}$ . For $n = 3$ , the best constant is about $0.47$ . I think for $n = 2017$ , it would be very difficult to find the exact value of the best constant, as opposed to finding the bounds we have established above.

Thanks! Your observations helped me a lot!

Steven Jim - 3 years, 10 months ago

@Jon Haussmann

Oh please don't give the source of the problem. I wanted to hide that.

Priyanshu Mishra - 3 years, 10 months ago

It's in the solution part. Don't worry.

Steven Jim - 3 years, 10 months ago

Telescopic Inequality!. Right?

The answer is 0.444.

1 solution

0 pending reports