Inequality $\frac{1}{a+b}+\frac{1}{a+2b}+...+\frac{1}{a+nb}<\frac{n}{\sqrt{a\left( a+nb \right)}}$ 
Let $a,b\in \mathbb{R+}$ and $n\in \mathbb{N}$. Prove that:
  $$\frac{1}{a+b}+\frac{1}{a+2b}+...+\frac{1}{a+nb}<\frac{n}{\sqrt{a\left( a+nb \right)}}$$

I have a solution using induction, but without induction I can not think of any way to prove this.
 A: Here is a proof that is an alternative to and perhaps more straightforward than the ingenious one of Michael Rozenberg.
WOLOG, let $a = 1$.
Since
$$1+t<\Big(1+\frac t2\Big)^2,$$
$$\sum_{i=1}^n \frac b{1+bi}< \int_0^{bn}\frac1{1+t}\,dt<\int_0^{bn}\frac{1+\frac t2}{(1+t)^\frac32}\,dt =\frac{t}{\sqrt{1+t}}\bigg|_{t=0}^{bn}.$$
QED
Indeed, from the derivation above, the left hand side of the original inequality has a tighter upper bound of $\frac1b\ln\big(1+\frac ba n\big)$.
A: For positives $a$ and $b$ By C-S we obtain:
$$\left(\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb}\right)^2$$
$$\leq n\left(\frac{1}{(a+b)^2}+\frac{1}{(a+2b)^2}+\cdots+\frac{1}{(a+nb)^2}\right)$$
$$<n\left(\frac{1}{(a+\frac{b}{2})(a+\frac{3b}{2})}+\frac{1}{(a+\frac{3b}{2})(a+\frac{5b}{2})}+\cdots+\frac{1}{(a+\frac{(2n-1)b}{2})(a+\frac{(2n+1)b}{2})}\right)$$
$$=\frac{n}{b}\left(\frac{1}{a+\frac{b}{2}}-\frac{1}{a+\frac{3b}{2}}+\frac{1}{a+\frac{3b}{2}}-\frac{1}{a+\frac{5b}{2}}+\cdots+\frac{1}{a+\frac{(2n-1)b}{2}}-\frac{1}{a+\frac{(2n+1)b}{2}}\right)$$
$$=\frac{n}{b}\left(\frac{1}{a+\frac{b}{2}}-\frac{1}{a+\frac{(2n+1)b}{2}}\right)=\frac{n^2}{(a+\frac{b}{2})(a+\frac{(2n+1)b}{2})}<\frac{n^2}{a(a+nb)}.$$
Done!
