Induction Inequality with Summation

Question

I can't seem to figure out this problem. Do you have any ideas?

For an integer $n > 1$, show that $$ \sum_{k=1}^n {1\over \sqrt{{n^2}+k}} > {{\sqrt{1+{1\over n}}}\over 2} $$

Answer 1

Notice first that $\sqrt{1+\frac1n}=\sqrt{\frac{n+1}{n}}=\frac{\sqrt{n^2+n}}{n}$, and therefore your inequality is equivalent to: $$\tag{1} \sum_{k=1}^n\frac{2n}{\sqrt{n^2+k}}>\sqrt{n^2+n}. $$ For every $k=1,2,\ldots,n$ we have $n^2+k\le n^2+n$, and therefore $$ \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}\ge \sum_{k=1}^n\frac{1}{\sqrt{n^2+n}}=\frac{n}{\sqrt{n^2+n}}. $$ It follows that $$ \sum_{k=1}^n\frac{2n}{\sqrt{n^2+k}}\ge \frac{2n^2}{\sqrt{n^2+n}}=\frac{2n^2}{n^2+n}\sqrt{n^2+n}=\frac{2n}{n+1}\sqrt{n^2+n}\equiv u(n)\sqrt{n^2+n}. $$ Now, the function
$$ u:[0,\infty)\to [0,\infty), u(t)=\frac{2t}{t+1}=2\left(1-\frac{1}{t+1}\right) $$ is differentiable, and $$ u'(t)=\frac{2}{(t+1)^2}>0, $$ i.e. $u$ is increasing. Thus $$ u(n)>u(1)=1\quad \forall n\ge 2, $$ and (1) follows.

Answer 2

An idea:

$$\sum_{k=1}^n\frac1{\sqrt{n^2+k}}\ge\frac n{\sqrt{n^2+n}}=\frac1{\sqrt{1+\frac1n}}\ge\frac{\sqrt{1+\frac1n}}2\iff 1+\frac1n\stackrel{\color{green}\checkmark}\le 2$$

Answer 3

Third proof: $f(x) = \dfrac{n}{\sqrt{x+n^2}}$ has $f''(x) = \dfrac{3n(x+n^2)^{-\frac{5}{2}}}{4}> 0$, thus $f$ is convex. Apply Jensen's inequality: $\displaystyle \sum_{k=1}^n \dfrac{n}{\sqrt{k+n^2}}=\displaystyle \sum_{k=1}^n f(k)\geq nf\left(\dfrac{1+2+\cdots+n}{n}\right)=n\cdot \dfrac{n}{\sqrt{n^2+\dfrac{1+2+\cdots + n}{n}}}=\dfrac{n^2}{\sqrt{n^2+\dfrac{n(n+1)}{2n}}} > \dfrac{\sqrt{n^2+n}}{2}\iff 6n^3>3n^2+2n+1$, and this is true by induction on $n$.