Proof for Inequality Can somebody tell me what is the name of the inequality:
\begin{equation} \sum_{t=1}^T \frac{1}{\sqrt{t}} \leq 2\sqrt{T}   \end{equation}
or any hint/link how to prove above? 
Thanks.

We will prove that for every $n \in \mathbb Z_{\gt 0}$ we have 
\begin{equation*} \tag{1}
\sum_{j=1}^n \frac{1}{\sqrt{j}} \lt 2 \sqrt{n}. 
\end{equation*}
We proceed by Mathematical Induction. 
Base case. The inequality is true for $n=1$ since $1 < 2$. 
Inductive step. Let $k \in  \mathbb Z_{\gt 0}$ be arbitrary. Assume that 
\begin{equation*} \tag{2}
\sum_{j=1}^k \frac{1}{\sqrt{j}} \lt 2 \sqrt{k}. 
\end{equation*} 
This is the inductive hypothesis. We need to prove that 
\begin{equation*} \tag{3}
\sum_{j=1}^{k+1} \frac{1}{\sqrt{j}} \lt 2 \sqrt{k+1}. 
\end{equation*}
Before proceeding with the proof of (3), we notice that 
$4k(k+1) < (2 k + 1)^2.$ Consequently, $2 \sqrt{k}\sqrt{k+1} < 2 k + 1$, and therefore, 
\begin{equation*} \tag{4}
2 \sqrt{k}\sqrt{k+1} + 1 < 2(k + 1). 
\end{equation*}
Now we proceed with the proof of (3). 
\begin{alignat*}{2} 
\sum_{j=1}^{k+1} \frac{1}{\sqrt{j}} & \lt 2 \sqrt{k} + \frac{1}{\sqrt{k+1}} & \qquad &  \text{by (2)}  \\
& = \frac{2 \sqrt{k} \sqrt{k+1} + 1}{\sqrt{k+1}} & & \\
& \lt \frac{2 (k+1)}{\sqrt{k+1}} & &  \text{by (4)} \\
& = 2 \sqrt{k+1}. & & 
\end{alignat*}
This proves that (2) implies (3) and completes the proof. 
