極值問題
設[img]http://latex.codecogs.com/gif.latex?x_{i}\geq 0(i=1,2,...,n)[/img]，且[img]http://latex.codecogs.com/gif.latex?\sum_{i=1}^{n}x_{i}^{2}+2\sum_{1\leq k%3Cj\leq n}^{}}\sqrt{\frac{k}{j}}x_{k}x_{j[/img]=1，求[img]http://latex.codecogs.com/gif.latex?\sum_{i=1}^{n}x_{i}[/img]之最大值? 令\(x_k=\sqrt{k}y_k\)
則\(\displaystyle \sum_{k=1}^n ky_k^2+2 \sum_{1 \le k<j \le n}ky_k y_j=1\)
令\(\cases{y_1+y_2+\ldots+y_n=a_1 \cr y_2+\ldots+y_n=a_2 \cr \ldots \cr y_n=a_n}\)
則\(a_1^2+a_2^2+\ldots+a_n^2=1\)
令\(a_{n+1}=0\)
則
\(\displaystyle \sum_{i=1}^n x_i\)
\(\displaystyle =\sum_{k=1}^n x_k\)
\(\displaystyle =\sum_{k=1}^n \sqrt{k}y_k\)
\(\displaystyle =\sum_{k=1}^n \sqrt{k}(a_k-a_{k+1})\)
\(\displaystyle =\sum_{k=1}^n \sqrt{k}a_k-\sum_{k=1}^n \sqrt{k}a_{k+1}\)
\(\displaystyle =\sum_{k=1}^n \sqrt{k}a_k-\sum_{k=1}^n \sqrt{k-1}a_k\)
\(\displaystyle =\sum_{k=1}^n (\sqrt{k}-\sqrt{k-1})a_k\)
\(\displaystyle \le \sqrt{\sum_{k=1}^n (\sqrt{k}-\sqrt{k-1})^2}\times \sqrt{\sum_{k=1}^n a_k^2}\)
\(\displaystyle =\sqrt{\sum_{k=1}^n (\sqrt{k}-\sqrt{k-1})^2}\)
回復 2# thepiano 的帖子
wow!十分感謝!
在"令\(a_{n+1}=0\)"
難怪書上把這題歸類在"阿貝爾變換"...
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