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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: http://www.artofproblemsolving.com/Wiki/index.php/Power_Mean_Inequality

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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.

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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.

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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
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