Here's the translated official solution. (I did not find this on my own.) Solution. For real numbers $x \geq \frac{1}{2}$, define \[ S(x) = \sum_{k=1}^{\left\lfloor{x}\right\rfloor} a_k \text{ and } T(x) = \sum_{k=1}^{\lfloor{x}\rfloor} \left\lfloor{\frac{x}{k}}\right\rfloor a_k. \] Claim. For any $\lambda \geq \frac{1}{2} $, we have \[S(2\lambda) - S(\lambda) \leq T(2 \lambda) -2T(\lambda). \tag{1}\] Proof. Obviously $S(x) = T(x) = 0$ for $\frac{1}{2} \leq x < 1$. Note that $\left\lfloor{\frac{2\lambda}{k}}\right\rfloor=1$ for $k \in \{ \lfloor{\lambda}\rfloor + 1, \dots, \lfloor{2\lambda}\rfloor\}$, so \begin{align*} S(2\lambda) - S(\lambda) &= \sum_{k= \lfloor{\lambda}\rfloor+1}^{\lfloor{2\lambda}\rfloor} = \sum_{k = \lfloor{\lambda} \rfloor+1}^{2\lfloor{\lambda}\rfloor} \left\lfloor{\frac{2\lambda}{k}}\right\rfloor a_k \\ &\leq \sum_{k=1}^{\lfloor\lambda\rfloor} \left( \left\lfloor{\frac{2\lambda}{k}}\right\rfloor - 2 \left\lfloor{\frac{\lambda}{k}}\right\rfloor \right) + \sum_{k = \lfloor\lambda\rfloor + 1}^{\lfloor{2\lambda}\rfloor} = T(2\lambda) - 2 T(\lambda) . \end{align*}The fact that $\sum_{k=1}^{\lfloor\lambda\rfloor} \left( \left\lfloor{\frac{2\lambda}{k}} \right\rfloor- 2 \left\lfloor{\frac{\lambda}{k}}\right\rfloor \right)$ stems from the fact that $\left\lfloor{\frac{2\lambda}{k}} \right\rfloor> 2\left\lfloor{\frac{\lambda}{k}}\right\rfloor$.This establishes the claim. Consider a real number $y \geq 1$ and choose the integer $m$ such that $2^{m-1} \leq y < 2^m$. Using (1), it can be seen that \begin{align*} S(y) &= S(y) - S\left(\frac{y}{2^m}\right) = \sum_{k=1}^m \left[ S\left(\frac{y}{2^{k-1}}\right) - S \left( \frac{y}{2^k} \right) \right] \\ &\leq \sum_{k=1}^m \left[ T\left( \frac{y}{2^{k-1}}\right) - 2T\left(\frac{y}{2^k}\right) \right] \\ &T(y) - \left[\sum_{k=1}^{m-1} T\left(\frac{y}{2^k}\right)\right] - 2T \left(\frac{y}{2^m} \right) = T(y) - \sum_{k=1}^{m-1} T\left(\frac{y}{2^k}\right). \end{align*}By the condition, we have that $x \log x - Cx < T(x) < x \log x + Cx$. Therefore \begin{align*} &S(y) \leq T(y) - \sum_{k=1}^{m-1} T\left( \frac{y}{2^k} \right) \leq y \log y + Cy - \sum_{k=1}^{m-1} \left(\frac{y}{2^k} \log \frac{y}{2^k} - \frac{Cy}{2^k} \right) \\ &= y \log y + Cy - \sum_{k=1}^{m-1} \frac{y}{2^k} \log y + \sum_{k=1}^{m-1 } \frac{y}{2^k} \log 2 + \sum_{k=1}^{m-1} \frac{Cy}{2^k} \\ &= y \log y\left(1- \sum_{k=1}^{m-1} \frac{1}{2^k} \right) + Cy\left( 1 + \sum_{k=1}^{m-1} \frac{1}{2^k} \right) + y \log 2 \sum_{k=1}^{m-1} \frac{k}{2^k} \\ &< \frac{1}{2^{m-1}} y \log y + 2Cy + 2 \log 2 \cdot y < 2 \log y + (2C + 2 \log 2) y. \end{align*}The penultimate step there uses the facts $1 + \sum_{k=1}^{m-1} 2^{-k} < 2$ and \[ \sum_{k=1}^{m-1} \frac{k}{2^k} = 2\sum_{k=1}^{m-1} \frac{k}{2^k} - \sum_{k=1}^{m-1} \frac{k}{2^k} = \sum_{k=0}^{m-2} \frac{k+1}{2^k} - \sum_{k=0}^{m-1} \frac{k}{2^k} = \left( \sum_{k=0}^{m-2} \frac{1}{2^k} \right) - \frac{m-1}{2^{m-1}} < 2. \]The last step comes from the fact that $1 \leq 2^{m-1} \leq y < 2^m$. Thus, to prove that $S(y) < 3Cy$, it is enough to show that $2 \log y + (2C + 2 \log 2)y < 3Cy$ , or equivalently \[ \frac{\log y}{y} < \frac{C}{2} - \log 2 . \]Let $\log y = t \geq 0$, so $y = 10^t = (1+9)^t \geq 1 +9t$; combined with $C \geq 1$, we see \[ \frac{\log y}{y} < \frac{1}{9} < \frac{1}{2} - \log 2 \leq \frac{C}{2} - \log 2 .\]Thus we conclude that $S(y) < 3Cy$ for all $y \geq 1$. Done.