<s>[INST] Proving by mathematical induction: $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>2(\sqrt{n+1}-1)$ 
Possible Duplicate:
Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction

I've tried to work on this equation for about 4-5 Hours. 
I'm trying to show that the following equation is true with induction:
$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>2(\sqrt{n+1}-1)$

after about 4-5 hours and numerous attempts I couldn't really get it to work out. 
Any ideas ?
 [/INST] There are ways other than induction (such as comparison with an integral) that have much clearer intuitive content. But let's stick with induction.  We need to deal with the induction step. Suppose that we know for a given $k$ that 
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{k}}>2(\sqrt{k+1}-1).\qquad\qquad(\ast)$$
We wish to show that 
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{k}}+   \frac{1}{\sqrt{k+1}}  >  2(\sqrt{k+2}-1).\qquad\qquad(\ast\ast)$$
By $(\ast)$, we have 
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{k}}+   \frac{1}{\sqrt{k+1}} >2(\sqrt{k+1}-1)+  \frac{1}{\sqrt{k+1}}.$$
We will be finished if we can prove that 
$$2(\sqrt{k+1}-1)+  \frac{1}{\sqrt{k+1}} >2(\sqrt{k+2}-1).$$
Equivalently, by a little algebra, we want to show that 
$$2\sqrt{k+1}+  \frac{1}{\sqrt{k+1}} >2\sqrt{k+2}.$$
Equivalently, by a little more algebra, we want to show that 
$$2(k+1)+1>2\sqrt{k+2}\sqrt{k+1}$$
(we multiplied throgh by the positive number $\sqrt{k+1}$).
So we want to show that
$$2k+3 >2\sqrt{k^2+3k+2}.$$
Since both sides are positive, this is equivalent to showing that
$$(2k+3)^2>4(k^2+3k+2).$$
Now we are finished, for the left side is $4k^2+12k+9$, while the right side is $4k^2+12k+8$, so the left side is greater than the right side.
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