<s>[INST] How prove this inequality $H(a_1)+H(a_2)+\cdots+H(a_m)\leq C\sqrt{\sum_{i=1}^{m}i a_i}$ 
Prove that: There exists a constant $C>0$ such that 
  $$H(a_1)+H(a_2)+\cdots+H(a_m)\leq C\sqrt{\sum_{i=1}^{m}i a_i}$$
  holds for arbitrary positive integer $m$ and any $m$ positive integers $a_1,a_2,\cdots,a_m$,
  where $H(n)=\sum_{k=1}^{n}\frac{1}{k}.$

This is 2018 China TST 3 Day 1 Q3. Maybe it is from some paper. The reason is that in the past, the Chinese training team selected most of the questions from articles.
 [/INST] In view of the rearrangement inequality, it suffices to check the inequality when $(a_i)$ is decreasing, i.e., $a_1 \geq a_2 \geq \cdots \geq a_m$. Also, it is no harm to introduce $a_{m+1} = 0$. Then
$$ \sum_{i=1}^{m} H(a_i)
= \sum_{i=1}^{m} \sum_{k=i}^{m} \left( H(a_k) - H(a_{k+1}) \right)
= \sum_{k=1}^{m} k \left( H(a_k) - H(a_{k+1}) \right) $$
and likewise
$$ \sum_{i=1}^{m} i a_i
= \sum_{i=1}^{m} \sum_{k=i}^{m} i (a_k - a_{k+1})
= \sum_{k=1}^{m} \frac{k(k+1)}{2} (a_k - a_{k+1}). $$
Now we invoke the following simple lemma:

Lemma. If $0 \leq a < b$ are integers, then
  $$ \frac{H(b) - H(a)}{\sqrt{b-a}}
\leq \sqrt{\frac{1}{a+\frac{1}{2}} - \frac{1}{b+\frac{1}{2}}} $$

Before proving this lemma, let us see how this implies the desired inequality. Applying the Cauchy-Schwarz inequality and the lemma above, we obtain
\begin{align*}
\sum_{i=1}^{m} H(a_i)
&\leq \left( \sum_{k=1}^{m} \frac{2\left( H(a_k) - H(a_{k+1}) \right)^2}{a_k - a_{k+1}} \mathbf{1}_{\{a_k > a_{k+1} \}} \right)^{1/2} \left( \sum_{k=1}^{m} \frac{k^2}{2} (a_k - a_{k+1}) \right)^{1/2} \\
&\leq \left( 2 \sum_{k=1}^{m-1} \left( \frac{1}{a_{k+1}+\frac{1}{2}} - \frac{1}{a_k+\frac{1}{2}} \right) \right)^{1/2} \left( \sum_{i=1}^{m} i a_i \right)^{1/2} \\
&\leq 2 \left( \sum_{i=1}^{m} i a_i \right)^{1/2}.
\end{align*}
Therefore the claim is true with $C = 2$.

Proof of Lemma. Notice that for $x \geq 1$,
$$
\int_{x-\frac{1}{2}}^{x+\frac{1}{2}} \frac{dt}{t}
= \int_{x}^{\infty} \left( \frac{1}{t-\frac{1}{2}} - \frac{1}{t+\frac{1}{2}} \right) \, dt
= \int_{x}^{\infty} \frac{4 dt}{4t^2-1}
\geq \int_{x}^{\infty} \frac{dt}{t^2}
= \frac{1}{x}.
$$
(Alternatively, this is the result of the convexity of $\frac{1}{x}$. Indeed, the tangent line $\frac{1}{x} - \frac{1}{x^2}(t-x)$ at $x$ lies below $\frac{1}{t}$, and integrating both sides from $x-\frac{1}{2}$ to $x+\frac{1}{2}$ gives the inequality above.) Then by Cauchy-Schwarz inequality,
\begin{align*}
H(b) - H(a)
&= \sum_{k=a+1}^{b} \frac{1}{k}
 \leq \int_{a+\frac{1}{2}}^{b+\frac{1}{2}} \frac{dx}{x} \\
&\leq \left( \int_{a+\frac{1}{2}}^{b+\frac{1}{2}}  \frac{dx}{x^2} \right)^{1/2}\left( \int_{a+\frac{1}{2}}^{b+\frac{1}{2}}  dx \right)^{1/2} \\
&= \Bigg( \frac{1}{a+\frac{1}{2}}-\frac{1}{b+\frac{1}{2}} \Bigg)^{1/2} \sqrt{b-a}.
\end{align*}
</s>