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.

  • $\begingroup$ 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. $\endgroup$ – Michal Adamaszek Oct 28 at 9:49
  • $\begingroup$ 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... $\endgroup$ – 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 $Y_i X_j - X_i Y_j$.


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