Track / Stage 6 / 84 of 400 #1084 of 1964

Problem 1084

National olympiad, first round

Algebra Difficulty 6.1 Prove it

Let $x_{1}, \ldots, x_{n}$ and $a_{1}, \ldots, a_{n}$ be real numbers in $[0,1]$ such that $\sum_{i=1}^{n} a_{i}=1$ . Show that for any $s \in \mathbb{R}$ ,

$\ln \left(\sum_{i=1}^{n} a_{i} e^{s x_{i}}\right) \leqslant\left(e^{s}-1\right) \sum_{i=1}^{n} a_{i} x_{i}$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Observe that for all $x \in [0,1]:$

$e^{s x}=e^{x s+(1-x) \cdot 0} \leqslant x e^{s}+(1-x) e^{0}=x e^{s}+(1-x)$

by the convexity of the exponential function. Thus:

$\begin{aligned} \sum_{i=1}^{n} a_{i} e^{s x_{i}} & \leqslant \sum_{i=1}^{n} a_{i}\left(x_{i} e^{s}+\left(1-x_{i}\right)\right) \\ & =e^{s} \sum_{i=1}^{n} a_{i} x_{i}+\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n} a_{i} x_{i} \\ & =1+\left(e^{s}-1\right) \sum_{i=1}^{n} a_{i} x_{i} \end{aligned}$

Since by hypothesis, $\sum_{i=1}^{n} a_{i}=1$ . Applying the inequality $1+x \leqslant e^{x}$ to this last expression:

$1+\left(e^{s}-1\right) \sum_{i=1}^{n} a_{i} x_{i} \leqslant \exp \left(\left(e^{s}-1\right) \sum_{i=1}^{n} a_{i} x_{i}\right)$

We obtain:

$\sum_{i=1}^{n} a_{i} e^{s x_{i}} \leqslant \exp \left(\left(e^{s}-1\right) \sum_{i=1}^{n} a_{i} x_{i}\right)$

The desired result is obtained by taking the logarithm of both sides of this inequality.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.