An inequality like Riemann sum involving $\sqrt{1-x^2}$ How can I prove that for every positive integer $n$ we have
\begin{equation*}
\frac{n\pi}{4}−\frac{1}{\sqrt{8n}}<\frac{1}{2}+\sum_{k=1}^{n−1}\sqrt{1−\frac{k^2}{n^2}}?
\end{equation*}

Write the inequality as
$$\frac{\pi}{4} < \frac{1}{2n} + \frac{1}{n} \sum_{k=1}^{n-1} \sqrt{1-\left(\frac{k}{n}\right)^2} + \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$
The left-hand side $\pi/4$ is the area of the part of the unit circle that
lies in the first quadrant (below the curve $y=f(x)=\sqrt{1-x^2}$).
We want to interpret the right-hand side as the area of a region $D$ which
covers that quarter circle.
Note that $f$ is concave, so that its graph lies below any tangent line.
Thus the trapezoid bounded by the lines $x=a-\epsilon$ and $x=a+\epsilon$
and by the $x$ axis and the tangent line through $(a,f(a))$ will cover the
corresponding part of the circle:
$$\int_{a-\epsilon}^{a+\epsilon} f(x) dx < 2\epsilon f(a).$$
Thus, taking $D$ to be the union of the following pieces does the trick:


*

*A rectangle of height 1 between $x=0$ and $x=1/2n$.

*Trapezoids as above, of width $\frac{1}{n}$ and centered at $x=k/n$ for $k=1,\ldots,n-1$.

*A trapezoid as above, of width $\frac{1}{2n}$ and centered at $x=1-1/4n$. This last one has area
$$\frac{1}{2n} f(1-1/4n) = \frac{1}{2n} \sqrt{\frac{1}{2n} - \frac{1}{16n^2}} <   \frac{1}{2n} \sqrt{\frac{1}{2n}}.$$

