Consider the generalized (power) mean of positive numbers $a_1, \dotsc, a_n$ $$M_p(a_1, \dotsc, a_n)=\left(\frac{a_1^p + \dotsb + a_n^p}{n}\right)^{1/p}\qquad p\in \mathbb{R}$$

where for $p=0$ we use the geometric mean. The generalized mean inequality says that $$ p < q \implies M_p(a_1, \dotsc, a_n) \leq M_q(a_1, \dotsc, a_n),$$

with equality holding iff $a_1 = \dotsb = a_n$. On Wikipedia it says that one can prove this by differentiating with respect to $p$ and using Jensen's inequality, and noting that $\partial_p M_p(a_1, \dotsc, a_n)>0$. When I do so, I get:

$$\partial_p \left(\left(\frac{a_1^p + \dotsb + a_n^p}{n}\right)^{1/p}\right)\\= \left(\frac{a_1^p + \dotsb + a_n^p}{n}\right)^{1/p} \partial _p \left( \frac1p \log (a_1^p + \dotsb + a_n^p) - \frac1p \log n \right),$$ which is positive iff $\partial _p \left( \frac1p \log (a_1^p + \dotsb + a_n^p) - \frac1p \log n \right)$ is positive.

$$\partial _p \left( \frac1p \log (a_1^p + \dotsb + a_n^p) - \frac1p \log n \right) \\= \left(\frac{-1}{p^2}\right) \log (a_1^p + \dotsb + a_n^p)+ \frac1p \frac{a_1^p \log a_1+ \dotsb + a_n^p \log a_n }{a_1^p + \dotsb + a_n^p}+\frac{1}{p^2}\log n.$$

I'm trying to show that this is positive. When I think about what Jensen's equality would tell me, I note that $$\frac{a_1^p\log a_1 + \dotsb + a_n^p \log a_n}{a_1^p + \dotsb + a_n^p} \\ \leq \log \left( \frac{a_1^p}{a_1^p + \dotsb + a_n^p}a_1 + \dotsb + \frac{a_n^p}{a_1^p + \dotsb + a_n^p}a_n \right) \\ = \log \left( \frac{a_1^{p+1} + \dotsb + a_n^{p+1}}{a_1^p + \dotsb + a_n^p} \right),$$ or we could try to bring the factor $\frac1p$ of the term in question into the mix and make a similar estimate.

Anyone have an idea?

  • 1
    $\begingroup$ I don't see why one would differentiate at all, Jensen's inequality directly yields the result. $\endgroup$ Commented Jan 22, 2014 at 20:08
  • $\begingroup$ @DanielFischer Wanting to treat $p>q>0$, $p>0>q$ and $0>p>q$ simultaneously would be one reason, although the computation is indeed ugly. $\endgroup$ Commented Jan 31, 2019 at 1:46

1 Answer 1


I agree with Daniel Fischer's comment: it's easier to use Jensen's directly (as here, of which the present question is a special case, since averages are a special case of integrals).

But maybe you really want to prove the derivative wrt $p$ is positive; after all, this gives more information than the function being strictly increasing. Then do the following: normalize $a_i$ (multiplying them by the same positive constant) so that $\sum a_i^p =1$. The scary inequality $$ \left(\frac{-1}{p^2}\right) \log (a_1^p + \dotsb + a_n^p)+ \frac1p \frac{a_1^p \log a_1+ \dotsb + a_n^p \log a_n }{a_1^p + \dotsb + a_n^p}+\frac{1}{p^2}\log n\ge 0$$ suddenly simplifies to $$ \frac1p (a_1^p \log a_1+ \dotsb + a_n^p \log a_n )+\frac{1}{p^2}\log n\ge 0$$ Writing $b_i=a_i^p$ simplifies it further: $$ \sum b_i \log b_i \ge \log (1/n) \tag{1}$$ with equality iff all $b_i$ are equal. You may have seen (1) in the context of entropy. Jensen's inequality yields (1) because the function $x\mapsto x\log x$ is convex on $[0,1]$.

x log x


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