Chứng minh giúp em ạ

Question

Cho

$a_1,a_2,...,a_n>0$ thỏa mãn:

$a_1+a_2+\cdots+a_n=1.$
Đặt

$H_k=\frac{k}{\displaystyle\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}}$ với

$k=1,2,\cdots, n$ .
Chứng minh rằng:

$H_1+H_2+\cdots+H_n<2.$

score 5 · Answer 1

Ta sẽ chứng minh:

$H_k\le\frac{4}{k(k+1)^2}(a_1+4a_2+\ldots+k^2a_k),\forall k=\overline{1,n}$

$(*)$
Thật vậy:

$\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_k}\right)(a_1+4a_2+\ldots+k^2a_k)$

$\ge\left(\sqrt{\frac{1}{a_1}}.\sqrt{a_1}+\sqrt{\frac{1}{a_2}}.2\sqrt{a_2}+\ldots+\sqrt{\frac{1}{a_k}}.k\sqrt{a_k}\right)^2$

$=(1+2+\ldots+k)^2=\frac{k^2(k+1)^2}{4}$ , suy ra

$(*)$ .
Từ đó, suy ra:

$H_1+H_2+\ldots+H_n\le x_1a_1+x_2a_2+\ldots+x_na_n$
với:

$x_i=4i^2\sum_{j=i}^n\frac{1}{j(j+1)^2},\forall i=\overline{1,n}$
Mà:

$\frac{1}{j(j+1)^2}<\frac{1}{2}.\frac{2j+1}{j^2(j+1)^2}=\frac{1}{2}\left(\frac{1}{j^2}-\frac{1}{(j+1)^2}\right)$
Suy ra:

$x_i<4i^2\sum_{j=i}^n\frac{1}{2}\left(\frac{1}{j^2}-\frac{1}{(j+1)^2}\right)$

$=2i^2\left(\frac{1}{i^2}-\frac{1}{(n+1)^2}\right)<2,\forall i=\overline{1,n}$

$\Rightarrow H_1+H_2+\ldots+H_n<2(a_1+a_2+\ldots+a_n)=2$

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