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For $x_1\ldots x_n \in \mathbb{R_{+}}$ show: $$\left(x_1x_2\ldots x_n\right)^{\frac{x_1+\cdots +x_n}{n}}\le x_1^{x_1}x_2^{x_2}\ldots x_n^{x_n}$$

Convexity and logarithmic function are the first ideas that came to me but I'd be interested in any other solutions. Consider $f:(0,\infty)\to \mathbb{R}$ with $x\mapsto x\ln x$

For $x>0$ we have $f''(x)=\frac{1}{x}>0$ thus $f$ is convex on $(0,\infty)$. By Jensen's inequality with the weights $p_i=1/n$, $i=1,\ldots ,n$

\begin{align} \ln \left(\frac{1}{n}\sum_{i=1}^{n} x_i \right)^{\frac{1}{n}\sum_{i=1}^{n} x_i} & \leq \frac{1}{n}\sum_{i=1}^{n} \ln x_i^{x_i} =\ln\left(x_1^{x_1}\ldots x_n^{x_n}\right)^{1/n} \\ \left( \frac{1}{n} \sum_{i} x_i\right)^{\sum_i x_i}&\leq \prod_{i} x_i^{x_i}\end{align}

So it suffices to show that \begin{align} \prod_{i}x_i^{\frac{1}{n}\sum_{i}x_i}& \leq \left( \frac{1}{n} \sum_{i} x_i\right)^{\sum_i x_i}\end{align}

Which could also possibly be achieved by convexity but I can't figure out how.

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    $\begingroup$ what you have in the end is just the regular AM-GM, after removing the common power $\sum_ix_i$ $\endgroup$
    – dezdichado
    Feb 8 at 18:14
  • $\begingroup$ embarrassing that I missed that, thank you $\endgroup$ Feb 8 at 18:53

1 Answer 1

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Another solution that may be simpler.

We observe that, for all $1\le i,j \le n$: $$(x_i - x_j)(\ln x_i - \ln x_j) \ge 0 \implies \color{red}{ x_i \ln x_j + x_j\ln x_i \le x_i \ln x_i + x_j\ln x_j} \tag{1}$$ Make the sum of $(1)$ over $1\le i,j \le n$: $$\begin{align} &\implies \sum_{1\le i,j \le n}\left(x_i \ln x_j + x_j\ln x_i \right) \le\sum_{1\le i,j \le n}\left(x_i \ln x_i + x_j\ln x_j \right) \\ &\iff2 \left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n \ln x_i\right) \le 2n \sum_{i=1}^n (x_i\ln x_i) \\ &\iff\frac{1}{n}\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n \ln x_i\right) \le \sum_{i=1}^n (x_i\ln x_i) \\ &\iff \left(x_1x_2\ldots x_n\right)^{\frac{x_1+\cdots +x_n}{n}}\le x_1^{x_1}x_2^{x_2}\ldots x_n^{x_n} \end{align}$$ Q.E.D

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