Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a 'simple' way to prove that $f\le F$ for any $\textit{n }$and for all $0<x_{i} ,y_{i} <1$?

I have only verified this numerically...

$$ \begin{array}{l} f(x_{1} ,...,x_{n} ,y_{1} ,...,y_{n} )= \\ {\frac{1}{n^{2} } \sum _{i=1}^{n}\left(2x_{i} (1-x_{i} )+2y_{i} (1-y_{i} )-4(-x_{i} ^{4} -y_{i} ^{4} +2x_{i} ^{3} +2y_{i} ^{3} -2x_{i} ^{2} -2y_{i} ^{2} +x_{i} +y_{i} )^{2} \right) } \end{array}$$

$$F=\frac{1}{n} \left(2x(1-x)+2y(1-y)-4(-x^{4} -y^{4} +2x^{3} +2y^{3} -2x^{2} -2y^{2} +x+y)^{2} \right) $$ where, $x=\frac{1}{n} \sum _{i=1}^{n}x_{i} ,\quad y=\frac{1}{n} \sum _{i=1}^{n}y_{i} $

share|cite|improve this question

I think you should try using Jensen's inequality, which says that if $g$ is a real convex function, then we have :

$g\left(\frac{1}{n}\sum x_i\right) \leq \frac{1}{n} \sum g(x_i)$

(and the inequality is reversed when $g$ is concave)

share|cite|improve this answer
Yes, I considered using Jensen's but g is a multivariate and I am not sure [a] how to use a multivariate version of Jensen's, and [b] how to determine if g is convex or concave... – Omri Feb 14 '12 at 15:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.