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)


You know that
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} \geq \sqrt{k}$$$$
and want to prove that:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k+1}} \geq \sqrt{k+1}$$$$
Add $\sqrt{k+1}-\sqrt{k}$ to both sides of first inequality, you get:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} +\sqrt{k+1}-\sqrt{k}\geq \sqrt{k+1}$$
But:
$$\sqrt{k+1}-\sqrt{k}=\frac{(\sqrt{k+1}+\sqrt{k})(\sqrt{k+1}-\sqrt{k})}{(\sqrt{k+1}+\sqrt{k})}=\frac{k+1-k}{(\sqrt{k+1}+\sqrt{k})}=\frac{1}{(\sqrt{k+1}+\sqrt{k})}\leq \\ \leq \frac{1}{\sqrt{k+1}}$$
So:
$$\sqrt{k+1} \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k}} +\sqrt{k+1}-\sqrt{k} \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{k+1}}$$
