Prove $\sum ^{n}_{k=1}\ln( k) \geq \sum ^{n}_{k=\left\lfloor \frac{n}{2}\right\rfloor +1}\ln( k)$

Question

hey I saw this inequalities in a soltion for some questions but I can't understand why this is true . Can somebody give me a hint about how they did that ? \begin{gather} \sum ^{n}_{k=1}\sqrt{k} \geq \sum ^{n}_{k=\left\lfloor \frac{n}{2}\right\rfloor +1}\sqrt{k}\\ \notag\\ \notag\\ \sum ^{n}_{k=1}\ln( k) \geq \sum ^{n}_{k=\left\lfloor \frac{n}{2}\right\rfloor +1}\ln( k) \end{gather}

Answer 1

The summand is nonnegative, so omitting terms cannot increase the sum: $$\sum_{k=1}^n f(k) = \sum_{k=1}^{\lfloor n/2\rfloor} f(k) + \sum_{k=\lfloor n/2\rfloor+1}^n f(k)\ge 0 + \sum_{k=\lfloor n/2\rfloor+1}^n f(k)$$