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

Most of the proofs of mean power inequality are based on jensen's inequality. Can the mean power inequality be prooved without use of one? Mean Power Inequality:

share|cite|improve this question
up vote 1 down vote accepted

It is hard to avoid Jensen, as the inequality does follow from convexity, but I'll give an elementary proof that doesn't use it explicitly.

Step 1: reduce the inequality to the case $k_2 = 1$. How is it done? Raise both sides of $M(k_1) \ge M(k_2)$ to the $k_2$'th power, and you'll get $M(\frac{k_1}{k_2}) \ge M(1)$, with $a_i' = a_i^{k_2}$.

Step 2: reduce the inequality to the case where $a_1 + \cdots + a_n = n$. It is "OK" to do so because $M(t)$ is homogeneous in the $a_i$'s.

Step 3: Lemma (the convexity argument) - $x+y = 2 \implies x^k+y^k \ge 2$ for any $k \ge 1$. Proof: let $x=1+\varepsilon, y=1-\varepsilon$ (WLOG $\varepsilon >0$). We then use Newton's binomial expansion: $$x^k+y^k = (1+\varepsilon)^k+(1-\varepsilon)^k = 2(\sum_{i=0}^{\infty} \varepsilon^{2i}\binom{k}{2i}) \ge 2$$ We even see that the inequality is strict unless $x=y=1$.

Step 4: We need to show that for any $k \ge 1: \sum_{i=1}^{n} a_i = n \implies \sum_{i=1}^{n} a_i^k \ge n$. The multivariate function $f(a_1,\cdots,a_n)=\sum a_i^k$ achieves a minimum on the compact set $\{a_i \ge 0 | \sum a_i = n\}$. We we'll show it occurs when $\forall i: a_i=1$. If, for contradiction's sake, we assume it occurs on $(a_1, \cdots ,a_n)$ when not all $a_i$'s are equal, we can replace it with the following better option: if WLOG $a_1$ is maximal and $a_n$ is minimal, we can replace them with their average: $$a_n' = a_1' = \frac{a_1+a_n}{2}$$ and we get a smaller value of $f$ - $$f(a_1,\cdots,a_n)-f(a_1',\cdots,a_n') = a_1^k + a_n^k - 2(\frac{a_1+a_n}{2})^k $$ $$= (\frac{a_1+a_n}{2})^k ((\frac{2a_1}{a_1+a_2})^k + (\frac{2a_2}{a_1+a_2})^k - 2) $$ and this is positive (by our lemma), giving a contradiction.

share|cite|improve this answer

As stated there, $$ \left(\frac1n\sum_{i=1}^na_i^{k_1}\right)^{1/k_1}\ge\left(\frac1n\sum_{i=1}^na_i^{k_2}\right)^{1/k_2}\tag{1} $$ when $k_1\ge k_2$ is almost directly Jensen's Inequality since $x^{k_1/k_2}$ is a convex function.

However, Hölder's inequality yields $$ \begin{align} \sum_{i=1}^na_i^{k_2}1 &\le\left(\sum_{i=1}^n\left(a_i^{k_2}\right)^{k_1/k_2}\right)^{k_2/k_1}\left(\sum_{i=1}^n1^{k_1/(k_1-k_2)}\right)^{(k_1-k_2)/k_1}\\ &=\left(\sum_{i=1}^na_i^{k_1}\right)^{k_2/k_1}n^{(k_1-k_2)/k_1}\tag{2} \end{align} $$ Which leads to $(1)$ after some simple algebra.

share|cite|improve this answer
Note that Hölder's inequality can be proven using Jensen's inequality, but not vice-versa, so Hölder is essentially weaker than Jensen. – robjohn Nov 23 '12 at 22:20
Thanks! That was really helpful. – 007resu Nov 23 '12 at 22:31

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.