Minimizing sum with constraints

Question

Minimize the following sum

$$\sum_{j=1}^n\frac{1}{x_j+1} $$ with constraints: for every $j, x_j\geq 0$ and $\sum_{j=1}^n x_j = 2m, m\in\mathbb{N}$

Intuitively, it makes sense to pick $x_j\equiv 2m/n$, for we we have to keep the denominator as large as possible for every summand. How do we justify this, though?

I could view the sum as $f(x_1, x_2, \ldots , x_n) = \sum_{j=1}^n\frac{1}{x_j+1}$ and now I need to determine a stationary point of this function i.e a point $(x_1', x_2',\ldots ,x_n')=:P$ with $\frac{\partial }{\partial x_j}f(x_j')\equiv 0$

..but $\frac{\partial f}{\partial x_j} \equiv -\frac{1}{(x_j+1)^2}$ which isn't zero for any $x_j'$.

Eventually I would like to apply the theorem which says if the Hessian is positive definite at a stationary point, then at that point, the function has a local minimum.

Answer 1

The approach with the Hessian would only work if the function as a global function $(\mathbb{R}^+)^n \to \mathbb{R}$ would have a local minimum and this minimum "coincidentially" satsifes the condition $\sum x_j = 2m$.

One possible approach would be to use lagrange multipliers, but there is an even simpler solution in this case.

From the AM-GM inequlity we know that $$\sqrt[n]{\prod \limits_{i = 1}^n x_i} \le \frac{1}{n} \sum \limits_{i = 1}^n x_i$$ holds for all positive $x_i$, where equality holds if and only if all the $x_i$ are the same. Applying this inequality to $\frac{1}{x_i}$ we get $$\sqrt[n]{\prod \limits_{i = 1}^n\frac{1}{x_i}} \le \frac{1}{n} \sum \limits_{i = 1}^n \frac{1}{x_i},$$ where again equality holds if and only if all the $x_i$ are the same. From the last inequality, we can conclude

the inequality of the harmonic, geometric and arithmetic mean: $$\frac{n}{\sum \limits_{i = 1}^n \frac{1}{x_i}} \le \sqrt[n]{\prod \limits_{i = 1}^n x_i} \le \frac{1}{n} \sum \limits_{i = 1}^n x_i,$$ where equality holds if and only if all the $x_i$ are the same.

If we apply the last inequality to $\frac{1}{x_i + 1}$, we get the following: $$\sum \limits_{i = 1}^n \frac{1}{x_i + 1} \ge \frac{n^2}{\sum \limits_{i = 1}^n (x_i + 1)} = \frac{n^2}{n + 2m},$$ where inequality holds if and only if all $x_i$ are the same, i.e. $x_i = \frac{2m}{n}$.