Question
Prove that there exists a constant $C > 0$ such that, for any integers $m, n$ with $n \geq m > 1$ and any real number $x > 1$,$$\sum_{k=m}^{n}\sqrt[k]{x} \leq C\bigg(\frac{m^2 \cdot \sqrt[m-1]{x}}{\log{x}} + n\bigg)$$
my attempt(wothless)
Lemma 1: For any real number $x > 1$, $$\sqrt[k]{x} \leq 1 + \frac{\log{x}}{k}$$
Lemma 2: For any positive integers $m$ and $n$, $$\frac{m^2}{2} \leq \sum_{k=m}^{n} \frac{1}{k}$$
Proof of Lemma 1:
By AM-GM, $$\sqrt[k]{x} = \left( x^{1/k} \right)^k \leq \frac{1}{k} \left( x + \frac{1}{k} + \frac{1}{k} + \dots + \frac{1}{k} \right) = 1 + \frac{\log{x}}{k}$$
Proof of Lemma 2:
By AM-GM, $$\frac{m + m + \dots + m}{n} \leq \left( \frac{m}{n} \right)^{n}$$
Multiplying both sides by $\frac{n}{2}$, $$\frac{m^2}{2} \leq \sum_{k=m}^{n} \frac{1}{k}$$
\begin{align*} \sum_{k=m}^{n} \sqrt[k]{x} &\leq \sum_{k=m}^{n} \left( 1 + \frac{\log{x}}{k} \right) \\ &= n + \sum_{k=m}^{n} \frac{\log{x}}{k} \\ &= n + \log{x} \cdot \sum_{k=m}^{n} \frac{1}{k} \\ &\leq n + \log{x} \cdot \frac{m^2}{2} \\ &= C \left( \frac{m^2 \sqrt[m-1]{x}}{\log{x}} + n \right) \end{align*}