Maximizing Reciprocals

This question was in Romania 1999. The answer is 1, which we shall prove using Olympiad inequalities.

First, we prove a lemma:

The maximum of the sum occurs when $n-1$ of the $x_i$ terms are equal. Consider $f(y)=\frac{1}{k+e^y}$ for an arbitrary nonnegative constant $k$ . We have $f'(y)=\frac {-e^y}{(k+e^y)^2}$ and $f''(y)=\frac{e^y (e^y-k)}{(k+e^y)^3}$ . Evidently, $f''(y) \geq 0 \iff e^y \geq k$ . Hence, $f(y)$ has a single inflexion point where $y=\ln(k)$ ., where $f(y)$ is convex over the interval $(\ln(k),\infty)$ . Now, we employ the substitution $y_i = \ln(x_i)$ so that $y_1 + \ldots + y_n = 0$ and

$\displaystyle \sum_{i=1}^n \dfrac {1}{n-1+x_i} = \displaystyle \sum_{i=1}^n f(y_i)$

We take $k=n-1$ and write $k_0 = \ln(n-1)$ . Suppose that $y_1 \geq \ldots \geq y_m \geq k_0 \geq y_{m+1} \geq \ldots \geq x_n$ for some positive $m$ . Then, by Majorisation,

$f(y_1) + \ldots + f(y_m) \leq (m-1)f(k_0) + f(y_1 + \ldots + y_m-(m-1)k_0)$

But then, also by Majorisation,

$(m-1)f(k_0)+f(y_{m+1})+\ldots+f(y_n) \leq (n-1)f(\dfrac{(m-1)k_0+y_{m+1}+\ldots+y_n}{n-1}$

Otherwise, all of the $y_i$ are less than $k_0$ . In that case, we may directly apply Majorisation to equate $n-1$ of the $y_i$ whilst increasing the sum in question. Hence, the lemma is valid.

Applying this lemma, it would suffice to show

$\dfrac {k}{k+x}+\dfrac {1}{k+\frac{1}{x^k}} \leq 1$

Clearing the denominators,

$\left (k^2 + \dfrac {k}{x^k} \right ) + (k + x) \leq k^2+ k \left (x+\dfrac{1}{x^k} \right ) + x^{1-k}$

$-xk+x+k \leq x^{1-k}$

But now this is evident. We have Bernoulli's inequality, since $x^{1-k} = (1+(x-1))^{1-k} \geq 1+(x-1)(1-k) = x+k-xk$ . Equality holds only where $x=1$ or $n=2$ .