<s>[INST] Prove lower bound $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+k^2}}\ge\left(1-\frac{1}{n}\right)\ln{(1+\sqrt{2})}+\frac{\sqrt{2}}{2n}$ 
Consider $$S_n=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k^2}}$$ and show that, for every positive integer $n$,
  $$S_n\ge\left(1-\frac{1}{n}\right)\ln{(1+\sqrt{2})}+\dfrac{\sqrt{2}}{2n}$$

I can prove a related upper bound:
$$S_n=\sum_{k=1}^{n}\dfrac{1}{n\sqrt{1+\left(\dfrac{k}{n}\right)^2}}\le\int_{0}^{1}\dfrac{1}{\sqrt{1+x^2}}dx=\ln{(1+\sqrt{2})}$$
but I can't prove the lower bound.
 [/INST] The inequality to be proved is 
$$\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+k^2}}>\left(1-\frac{1}{n}\right)\ln{(1+\sqrt{2})}+\frac{\sqrt{2}}{2n}$$
equivalent to
$$\frac{1}{n-1}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{1+(\frac{k}{n})^2}}>\ln{(1+\sqrt{2})}=\int_0^1\frac{1}{\sqrt{1+x^2}}.$$
LHS is the Riemann Sum of integral $\displaystyle\int_0^1\frac{1}{\sqrt{1+x^2}}$. When $n\to\infty$, LHS=RHS.
So what we have to prove is that LHS is monotonically decreasing with $n$.
Consider a funtion $f(x)=\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+(1-x)^2}}$.
Easy to know its 2nd derivative in $[0,1]$ is negative. 
So we have
$$f\Big(\frac{k}{n}\Big) > \Big(1-{k\over n}\Big)f\Big({k\over n+1}\Big)+{k\over n}f\Big({k+1\over n+1}\Big).$$ 
Sum over $n$, we get
$$\sum_{k=1}^{n-1} f\Big(\frac{k}{n}\Big)>\sum_{k=1}^{n-1}\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\sum_{k=1}^{n-1}\frac{k}{n}f\Big(\frac{k+1}{n+1}\Big)=\frac{n-1}{n}\sum_{k=1}^nf\Big(\frac{k}{n+1}\Big).$$
So $$\frac{1}{n-1}\sum_{k=1}^{n-1}\frac{1}{\sqrt{1+\Big(\dfrac{k}{n}\Big)^2}}=\frac{1}{2(n-1)}\sum_{k=1}^{n-1} f\Big(\frac{k}{n}\Big)$$ is monotonically decreasing with $n$. Q.E.D.
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