How prove this inequality $H(a_1)+H(a_2)+\cdots+H(a_m)\leq C\sqrt{\sum_{i=1}^{m}i a_i}$

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 $C = 2$.

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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*}

Sketch of an almost proof since I'm on my phone.

$H(a_i)<\ln(a_i+1) < ca_i^{1/2}$ for some $c>0$.

Therefore $\sum H(a_i) <\sum ca_i^{1/2} $.

Since $\frac1{m}\sum a_i^{1/2} \le \sqrt{\frac1{m}\sum a_i}$ by the power mean inequality, $\sum a_i^{1/2} \le \sqrt{m\sum a_i}$ so that $\sum H(a_i) <c\sqrt{m\sum a_i}$.

So if we can show that $m\sum a_i < b \sum ia_i $ for some $b$ we are done.

This is true if we can choose $b=m+1$ but not if $b$ is independent of $m$. For example, choose $a_1$ large ($m^2$) and all the others small (1). The left side is about $m^3$ and the right side is about $bm^2$.

I don't know where to go from here so I'll stop.