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.

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*}
