<s>[INST] Mathematical Induction Question, Proof Help Prove using Mathematical Induction that for all natural numbers ($n>0$):
$$
\frac 1 {\sqrt{1}} + \frac 1 {\sqrt{2}} + \cdots + \frac 1 {\sqrt{n}} \ge \sqrt{n}.
$$   
 Proof by Induction:
         Let P(n) denote    1/ √1 + 1/ √2 +  … + 1/ √n ≥  √n
 Base Case: n = 1, P(1) = 1/√1 ≥ √1
  The base cases holds true for this case since the inequality for P(1) holds true.

  Inductive Hypothesis: For every n = k > 0 for some integer k 
  P(k) = 1/ √1 + 1/ √2 +  … + 1/ √k ≥  √k, p(k) holds true for any integer k

 Inductive Step:
          P(k + 1)) = 1/ √1 + 1/ √2 +  … + 1/ √k  + 1/ √(k + 1) ≥  √k + √(k+1)
         √k + √(k+1) > √(k+1)   (this is where I got stuck)

 [/INST] Assuming $\sum_{i=1}^n \frac{1}{\sqrt{i}} \geq \sqrt{n}$, we have
\begin{align*}
\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} &= \sum_{i=1}^n \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{n+1}} \\
&\geq \sqrt{n} + \frac{1}{\sqrt{n+1}} \\
&= \frac{\sqrt{n^2+n} + 1}{\sqrt{n+1}} \\
&\geq \frac{\sqrt{n^2} + 1}{\sqrt{n+1}} = \frac{n+1}{\sqrt{n+1}} = \sqrt{n+1}.
\end{align*}
</s>