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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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