Mathematical Induction Proof: $1/\sqrt1 + 1/\sqrt2+...+1\sqrt n \ge \sqrt n$

Question

I need the full proof/solution to this problem.

What I've done so far:

Answer 1

suppose your formula is true for $n$, Then $$\sum^{n}_{i=1}{\frac{1}{\sqrt{i}}}>\sqrt{n}$$ we then try to prove the formula true for $n+1$ $$\sum^{n}_{i=1}{\frac{1}{\sqrt{i}}}+\frac{1}{\sqrt{n+1}}>\sqrt{n}+\frac{1}{\sqrt{n+1}}$$ $$=\frac{\sqrt{n}\sqrt{n+1}+1}{\sqrt{n+1}}>\frac{n+1}{\sqrt{n+1}}=\sqrt{n+1}$$

The base case is true so it's true for all n

Answer 2

Hint: note that $$ \frac 1{\sqrt n} \geq \sqrt n - \sqrt{n-1} $$

Answer 3

Multiply at both sides by $\sqrt{n}$. You get

$\sum_{k=1}^n\frac{\sqrt{n}}{\sqrt{k}}=\sum_{k=1}^n\frac{n}{\sqrt{kn}}\geq \sum_{k=1}^n\frac{n}{\sqrt{nn}}=\sum_{k=1}^n1=n$

Answer 4

Your proof is basically OK except for near the end, where you write

$${\sqrt{k(k+1)}+1\over\sqrt{k+1}}={\sqrt{k+1}\times\sqrt{k+1}\over\sqrt{k+1}}$$

The "$=$" there should be a "$\ge$."