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

For the function $$f = \frac{n}{\sum_i 1/x_i}$$ which is actually the harmonic mean of the values $$x_1,x_2,\ldots,x_n$$. How can we prove this is concave function? 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 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 at 19:18

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