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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.

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1 Answer 1

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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Robert, thank you for your answer, but can you exaplain why the largest is that in your answer? Thanks. –  lovm Nov 15 '12 at 20:46
    
It'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 $$f\left(\frac{\lambda_1 + \ldots + \lambda_n}{n}, \ldots, \frac{\lambda_1 + \ldots + \lambda_n}{n}\right) \le f(\lambda_1, \ldots, \lambda_n)$$ –  Robert Israel Nov 15 '12 at 22:49
    
Dear Robert, just one thing, why does the maximum occur when all $\lambda_ix_i$ equal? Thanks a lot! –  lovm Dec 7 '12 at 0:00
    
If, 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. –  Robert Israel Dec 7 '12 at 0:16
    
Dear Robert, what you just said shows the local maximum occurs there, but why does the global maximum also occur there? Thanks a lot again! –  lovm Dec 7 '12 at 0:23

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