Home > Analysis, Inequalities, Problem Solving > Traian Lalescu contest 2009 Problem 3

## Traian Lalescu contest 2009 Problem 3

Suppose $n \in \Bbb{N}, \ n \geq 2$ and $x_1,...,x_n >0,\ x_1+x_2+...+x_n=1$.Prove that

$\displaystyle \sum_{k=1}^n \frac{x_k}{1+k(x_1^2+x_2^2+...+x_k^2)}< \frac{\pi}{4}$, and the constant $\pi/4$ is the best possible.

Traian Lalescu contest 2009

1. May 11, 2011 at 2:53 pm

I don’t have a solution for the inequality, but if one can show it then we can conclude $\pi/4$ is the optimal constant by picking $x_1 = x_2 = x_3 = ... = x_n = 1/n$, which makes the sum a Riemann integral for $\int_0^1 1/(1+x^2) dx = \pi/4$, so we can make the sum arbitrarily close to $\pi/4$ by picking sufficiently large $n$.

• May 11, 2011 at 6:29 pm

Thank you for your remark. If you noticed that, you are very close to solving the inequality. The idea is to apply Cauchy Schwarz and then see that the result is indeed a Riemann sum for the given integral, which is smaller than the integral itself.