Suppose that $\sum_{i=1}^{n}\lambda_{i}=1$, where $\lambda_{i}>0$, and $\sum_{i=1}^{n}x_{i}^{2}=1$, where $x_{i}>0$. Does one have $n^{3/2}\min_{1\le i\le n}\lambda_{i}x_{i}\le B$ for some constant $B$ (independent of $n$)? Thanks.
Is there a constant for this?
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real-analysis
analysis
inequality
multivariable-calculus
1 Answers
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The largest $\min_i \lambda_i x_i$ can be is when all $\lambda_i = 1/n$ and all $x_i = 1/\sqrt{n}$, which makes the left side $1$.
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0Robert, thank you for your answer, but can you exaplain why the largest is that in your answer? Thanks. – 2012-11-15
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0It's easy to see that to maximize the minimum, you want to make all $\lambda_i x_i$ equal. Now if all $\lambda_i x_i = t$ so $x_i = t/\lambda_i$, we have $1 = \sum_i x_i^2 = t^2 \sum_i 1/\lambda_i^2$. Since $f(\lambda_1,\ldots,\lambda_n) = \sum_i 1/\lambda_i^2$ is a convex function and symmetric in $\lambda_1, \ldots, \lambda_n$, by averaging over cyclic permutations we get $$f\left(\frac{\lambda_1 + \ldots + \lambda_n}{n}, \ldots, \frac{\lambda_1 + \ldots + \lambda_n}{n}\right) \le f(\lambda_1, \ldots, \lambda_n)$$ – 2012-11-15
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0Dear Robert, just one thing, why does the maximum occur when all $\lambda_ix_i$ equal? Thanks a lot! – 2012-12-07
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0If, say, $\lambda_1 x_1 < \lambda_2 x_2$, then you increase the minimum by increasing $\lambda_1$ and $x_1$ slightly while decreasing $\lambda_2$ and $x_2$ to keep $\lambda_1 + \lambda_2$ and $x_1^2 + x_2^2$ constant. – 2012-12-07
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0Dear Robert, what you just said shows the local maximum occurs there, but why does the global maximum also occur there? Thanks a lot again! – 2012-12-07
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0A continuous function on a compact set attains a maximum (to get a compact set, you have to change $>0$ to $\ge 0$, but the value is $0$ when some $\lambda_i$ or $x_i$ is $0$). – 2012-12-07