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

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.

• You can for example prove that the set of points below the graph of $f$ (that is $\{x\in\mathbb{R}^{n+1}~:~ x_{n+1}\leq f(x_1,\ldots,x_n) \}$ is a convex set. – Michal Adamaszek Oct 28 '19 at 9:49
• In the case $n=2$ the proof is pretty simple: we have $$\text{AM}(x_1,x_2)-\text{HM}(x_1,x_2) = \frac{(x_1-x_2)^2}{2(x_1+x_2)}$$ and the RHS is the product of two positive and convex functions ($z^2$ and $\frac{1}{2w}$) of the independent variables $z=(x_1-x_2)$ and $w=(x_1+x_2)$. On the other hand the extension to $n>2$ does not seem to be straightforward... – Jack D'Aurizio Oct 28 '19 at 19:18

## 2 Answers

$$\text{HM}(x_1,\ldots,x_n)$$ 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 $$Y_i X_j - X_i Y_j$$.

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.

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.

• I cannot follow how you obtained Hessian in the general form but the result is wrong because at least for $0<p<1$, the p norm is neither convex nor concave: math.stackexchange.com/questions/2819671/… – Piyush Singh Jan 20 at 11:12
• @Piyush right, it should say on the domain of positive numbers. – Thomas Ahle Jan 20 at 20:00