Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given that $\{\phi_n\}$ is a sequence of non-negative numbers whose sum is $1$ and $\{\psi_n\}$ is a sequence of positive numbers, how can I show that $$ \prod_{n=1}^{\infty}~\psi_n^{\phi_n}~\leq~\sum_{n=1}^{\infty}~\phi_n\psi_n~? $$

Thanks.

PS: I'm not too sure about the title. Perhaps, someone could give it a better title?

share|improve this question
add comment

3 Answers

up vote 12 down vote accepted

Here is the proof from page 92 of The Cauchy-Schwarz Master Class by J. Michael Steele.

Since the exponential function is convex, Jensen's inequality gives $$\exp\left(\sum_n \phi_n y_n\right)\leq \sum_n \phi_n e^{y_n}.$$

Letting $\psi_n=e^{y_n}$ above we get $$\prod_{n}\ \psi_n^{\phi_n} \leq \sum_n \phi_n \psi_n.$$

share|improve this answer
2  
Aaaah... Glad to see someone referring to this wonderful book. +1. –  Did Nov 12 '11 at 21:08
1  
@Didier I agree. It is one of my favorites! –  Byron Schmuland Nov 12 '11 at 21:11
    
(+1) To both you and Didier. The Steele book is another (somewhat) "hidden" gem that we share an affection for. :) –  cardinal Nov 12 '11 at 21:20
add comment

Your inequality $$\prod_{n=1}^{\infty}~\psi_n^{\phi_n}~\leq~\sum_{n=1}^{\infty}~\phi_n\psi_n$$ is equivalent to $$\sum_{n=1}^{\infty} {\phi_n} \ln \psi_n \leq \ln \left(\sum_{n=1}^{\infty} \phi_n\psi_n\right).$$

This can be obtained from the integral form of Jensen's inequality: The following is true for any concave function $$\varphi\left(\int_\Omega g\, d\mu\right) \ge \int_\Omega \varphi \circ g\, d\mu. $$ (I have copied it from wikipedia, just changed convex to concave and change the inequality sign.)

You obtain the above inequality for:

  • concave function $\varphi(x)=\ln(x)$
  • measure space $\Omega=\mathbb N$ with the measure determined by $\mu(\{n\})=\phi_n$
  • function $g:\Omega\to\mathbb R$ given by $g(n)=\psi_n$.

The measure $\mu$ is sometimes called discrete measure.

Or, equivalently, you can look at $\psi_n$'s as probabilities and then use the probabilistic form of Jensen's inequality. (Comparing the mean value of logarithm and logarithm of the mean value.)


If you prefer, you might also use the finite version of Jensen's inequality: $$\varphi\left(\frac{\sum_{i=1}^k a_i x_i}{\sum_{j=1}^k a_j}\right) \ge \frac{\sum_{i=1}^k a_i \varphi (x_i)}{\sum_{j=1}^k a_j}$$ (again - the above formula is copied from wikipedia and I changed the sign) and then take the limit.

Alternatively, you can use the finite version to prove $$\prod_{n=1}^{N} \psi_n^{\phi_n} \leq \sum_{n=1}^{N} \phi_n\psi_n \qquad (*)$$ for arbitrary finite sequence $(\phi_n)$ fulfilling $\sum_{n=1}^{N} \phi_n=1$, use this for sequence and then use this to derive the same thing for your original -- infinite -- sequence, by taking the inequality (*) for $\phi_1,\dots,\phi_N,1-\sum_{i=1}^N \phi_i$ and $\psi_1,\dots,\psi_n,0$. (In this way the limits are slightly less complicated and you do not have the sum $\sum_{j=1}^k \phi_j$ in denominator.)


BTW It's worth mentioning that this is a generalization of AM-GM inequality.

share|improve this answer
add comment

We can apply Jensen inequality for a finite sums, i.e. $\sum_{k=1}^N\alpha_kf(x_k)\leq f\left(\sum_{k=1}^n\alpha_kx_k\right)$ if $\sum_{k=1}^n\alpha_k=1$. To see that, fix $n\in\mathbb N$ and put $c_k:=\frac{a_k}{\sum_{j=1}^na_j}$. Then by concavity of $x\mapsto \ln x$: \begin{align} \ln\prod_{k=1}^nb_k^{a_k}&=\sum_{k=1}^na_k\ln b_k\\ &=\left(\sum_{j=1}^na_j\right)\sum_{k=1}^nc_k\ln b_k\\ &\leq \left(\sum_{j=1}^na_j\right)\ln\left(\sum_{k=1}^nc_k b_k\right) \\ &=\left(\sum_{j=1}^na_j\right)\left(\ln\left(\sum_{k=1}^na_k b_k\right)-\ln\left(\sum_{j=1}^na_j\right)\right)\\ &\leq\ln\left(\sum_{k=1}^{+\infty}a_k b_k\right)-\ln\left(\sum_{j=1}^na_j\right), \end{align} and since $\lim_{n\to\infty}\sum_{j=1}^na_j=1$ we get the result, taking the $\limsup$.

share|improve this answer
add comment

Your Answer

 
discard

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.