TITLE: Prove $\sum ^{n}_{k=1}\ln( k) \geq \sum ^{n}_{k=\left\lfloor \frac{n}{2}\right\rfloor +1}\ln( k)$
QUESTION [0 upvotes]: 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}

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