Can we prove this inequality $1+\sqrt{2}+\cdots+\sqrt{n}<\frac{4n+3}{6}\sqrt{n}$ using integrals by parts? Can we prove this inequality $1+\sqrt{2}+\cdots+\sqrt{n}<\dfrac{4n+3}{6}\sqrt{n}$ using integrals by parts?
I need to prove this inequality, and I tried this way:
$$\begin{split}
\int_0^n\sqrt{x} \,dx &=\sum\limits_{k=1}^n \int_{k-1}^k \sqrt{x}\, dx\\
&=\sum\limits_{k=1}^n \int_0^1 \sqrt{t+k-1} \, dt\\
&=\sum\limits_{k=1}^n \sqrt{k}-\sum\limits_{k=1}^n \int_0^1\dfrac{t}{2\sqrt{t+k-1}}\,dt
\end{split}$$
Thus 
$$\sum\limits_{k=1}^n \sqrt{k}=\int_0^n \sqrt{x}\, dx+\sum\limits_{k=1}^n \int_0^1 \dfrac{t}{2\sqrt{t+k-1}}\, dt$$.
But maybe I didn't find some inequality trick, I cannot find a proper approxmation of the sum of the integrals. 
 A: Hint: Prove that $$\int_{k-1}^k\sqrt{x}dx>\dfrac{\sqrt{k}+\sqrt{k-1}}{2}$$ and then sum up to get your desired conclusion. You can just do this manually or generally this follows from a very well-known fact that the trapeziodal rule underestimates integral if the function is concave down. ($f(x) = \sqrt{x}$) is concave down)
A: First, see that we have
$$
\int_0^n \sqrt{x} \;dx = \frac{4n}{6}\sqrt{n}
$$
which gives us a "part" of the demanded result. Moreover, it suggests that we should try to prove the following:
$$
\sum_{k=1}^n \int_0^1  \frac{t}{2\sqrt{t+k-1}} \;dt < \frac{3}{6}\sqrt{n} = \frac{1}{2}\sqrt{n}
$$
Evaluating the integral we get:
\begin{align}
\int_0^1  \frac{t}{2\sqrt{t+k-1}} \;dt &= 
\int_{k-1}^{k}  \frac{x-k+1}{2\sqrt{x}} \;dx =
\frac{1}{2} \int_{k-1}^{k}  \sqrt{x} \;dx - (k-1)
\int_{k-1}^{k}  \frac{1}{2\sqrt{x}} \;dx \\
&=
\frac{1}{3} \left(-2k\sqrt{k} + 2k\sqrt{k-1} + 3\sqrt{k} - 2\sqrt{k-1}\right)
\end{align}
However, for all $k \geq 1$, you can prove that:
$$
\frac{1}{3} \left(-2k\sqrt{k} + 2k\sqrt{k-1} + 3\sqrt{k} - 2\sqrt{k-1}\right) < 
\frac{1}{2} \left(\sqrt{k}-\sqrt{k-1}\right)
$$
So we end up with a telescoping sum and the result we wanted:
$$
\sum_{k=1}^n \int_0^1  \frac{t}{2\sqrt{t+k-1}} \;dt < 
\frac{1}{2} \sum_{k=1}^n \left(\sqrt{k}-\sqrt{k-1}\right) = 
\frac{1}{2} \sqrt{n}
$$
