Find the maximum value of $x_1x_2+x_2x_3+\cdots+x_nx_1$ if $x_1+x_2+\cdots+x_n=0$ and $x_1^2+x_2^2+...+x_n^2=1$.
I can get by Cauchy-Schwarz that $x_1x_2+x_2x_3+\cdots+x_nx_1 \leqslant 1$, but can expression $x_1x_2+x_2x_3+\cdots+x_nx_1$ be smaller?
Asked
Active
Viewed 261 times
5
-
Hint: you can use Lagrange multipliers. – Gribouillis Aug 22 '17 at 21:10
-
1It's a very hard Fan's inequality. – Michael Rozenberg Aug 22 '17 at 21:13
-
2The answer is $\cos\frac{2\pi}{n}$. Ky Fan got this result in 1955. – Michael Rozenberg Aug 22 '17 at 21:29
-
What is your expression when $n=4$, is it $x_1x_2+x_2x_3+x_3x_4+x_4x_1$, or something else? – HorizonsMaths Aug 23 '17 at 01:21
1 Answers
2
This is known as the Ky Fan-Tausski-Todd inequality or a special case of the discrete Wirtinger's inequality. The proof is not that hard, but requires an ingenious change of variables. See here for the complete proof and other details.
dezdichado
- 13,888