How can we prove the harmonic mean is a concave function?

Question

How can we prove that the following function (harmonic mean) is concave?

$$f (x_1, x_2, \dots, x_n) := \frac{n}{\displaystyle\sum_{i=1}^n \frac{1}{x_i}}$$

Because this is not a function of one variable, I am not sure how to do so.

Answer 1

$\text{HM}(x_1,\ldots,x_n)$ is quite blatantly a continuous function on $\mathbb{R}_+^n$, so its concavity follows from its midpoint-concavity, i.e. from the inequality

$$ \frac{1}{\sum_{k=1}^{n}\frac{1}{x_k+y_k}} \geq \frac{1}{\sum_{k=1}^{n}\frac{1}{x_k}}+\frac{1}{\sum_{k=1}^{n}\frac{1}{y_k}} $$ which is the super-additivity of the harmonic mean: see §14 here. Up to a change of variables, this is equivalent to the following inequality for positive variables $$ \sum X_k \sum Y_k \geq \sum (X_k+Y_k) \sum \frac{X_k Y_k}{(X_k + Y_k)} \tag{A}$$ which can be proved by applying Lagrange multipliers to the determination of $$ \max_{\substack{\sum X_k = A \\ \sum Y_k = B}}\sum \frac{X_k Y_k}{X_k + Y_k}.$$ Since $\frac{\partial}{\partial X_k}\left( \frac{X_k Y_k}{X_k + Y_k}\right) = \left(\frac{Y_k}{X_k+Y_k}\right)^2 $ the maximum $\frac{AB}{A+B}$ is achieved when $\{X_k\}=\lambda \{Y_k\}$. In a more elementary way, the difference between the LHS and the RHS of $(A)$ is given by the sum over $i\neq j$ of $\frac{1}{(X_i+Y_i)(X_j+Y_j)}$ times

$$\left[(X_i+Y_i)(X_j+Y_j)(X_i Y_j+X_j Y_i)\right]-\left[X_j Y_j(X_i+Y_i)^2+X_i Y_i(X_j+Y_j)^2\right]$$ which can be checked to be the square of $Y_i X_j - X_i Y_j$.

Answer 2

Here is a proof using the Hessian, inspired by https://math.stackexchange.com/a/452822/7072 . Let $$f(x)=\frac1{\sum_i x_i^{-1}},$$ Then the Hessian is given by $$D_{ij}f=2 f^3 A \quad \text{where } \ A_{ij}= \begin{cases}-x_i^{-3}\sum_{k\neq i}x_k^{-1} \quad &\text{ if }\ i=j \\ x_i^{-2}x_j^{-2} \quad &\text{ if }\ i\ne j \end{cases}$$ We are to prove that $v^TAv\le 0$ for every vector $v$. (We are assuming that $f(x)$ is non-negative.) We find $$v^TAv=\left(\sum_{i=1}^n \frac{v_i}{x_i^2}\right)^2- \sum_{i=1}^n \frac{v_i^2}{x_i^3}\sum_j\frac1{x_i},$$ which is $\le 0$ by the Cauchy-Schwarz inequality applied to $(\frac{v_i}{x_i^{3/2}})\cdot (\frac{1}{x_i^{1/2}}) $.

My understanding from http://groups.uni-paderborn.de/fg-qi/courses/CMSC691/2016/assignments/CMSC691_Fall2016_A2.pdf is that every power mean is concave for $p<1$. I suppose it makes sense the proofs are somewhat similar.

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I was curious about what happens in the general $p$ case. In this case $f(x) = \left(\sum_i x_i^p\right)^{1/p}$ and we get $$D_{ij}f=\frac1{1-p} f^{1-2p} A \quad \text{where } \ A_{ij}= \begin{cases}-x_i^{p-2}\sum_{k\neq i}x_k^p \quad &\text{ if }\ i=j \\ x_i^{p-1}x_j^{p-1} \quad &\text{ if }\ i\ne j \end{cases} $$ Again, we find $$v^TAv=\left(\sum_{i=1}^n v_i x_i^{p-1}\right)^2- \sum_{i=1}^n v_i^2 x_i^{p-2}\sum_jx_i^p,$$ which is $\le 0$ by the Cauchy-Schwarz inequality applied to $( v_i x^{p/2-1})\cdot (x_i^{p/2}) $.

Note the same proof shows convexity for $p>1$ when the factor $\frac1{1-p}$ is positive.

Answer 3

As already stated Thomas Ahle by the harmonic mean in this case is the same as taking the finite lp mean on $\mathbb{R}^n$ rescaled by $\frac1n$ with here is where the magic happens $p=-1$. $$f(x)=\frac1n(\sum_{i=1}^n \frac1n(x_i)^{-1})^{-1}$$ We can then finish this argument by proofing the concavity of this lp mean. For that we can generalize to the setting where we have a measure space $(\Omega,\mathbb{P})$ and define $H_\mathbb{P}(x)=(\int x^{-1} d\mathbb{P})^{-1}$ and $x$ is now a measurable non negative function. Then the proof of its concavity is the same as the proof that any $L_p$ norm for $p\geq1$ is convex. We apply first the Minkowski inequality for $p<1$ (198. in Hardy, G. H., Littlewood, J. E., and Pólya, G. Inequalities. Univ. Press, Cambridge, 2nd ed. edition, 1952) which is exactly the other way around in this case and applicable as long as $\int x^{-1} d\mathbb{P}$ is finite in the following, and then the homogenity. Thus let now $g,h$ be measurable nonnegative such that their reciprocals are $\mathbb{P}$ integrable and $\lambda\in(0,1)$ then $$H_\mathbb{P}(\lambda g + (1-\lambda)h)\geq H_\mathbb{P}(\lambda g) + H_\mathbb{P}((1-\lambda)h)=\lambda H_\mathbb{P}(g) + (1-\lambda)H_\mathbb{P}(h).$$ The direct proof without the generalization would use the reverse Minkowsik inequality in the finite case wich would be 24 in (Hardy, G. H., Littlewood, J. E., and Pólya, G. Inequalities. Univ. Press, Cambridge, 2nd ed. edition, 1952)