Getting bounds from Riemann sums $y=\sum_{i=0}^9\frac{1}{10\sqrt 3} \frac{1}{1+(\frac{i}{10\sqrt 3})^2}$ and $x=\sum_{i=1}^{10}\frac{1}{10\sqrt 3} \frac{1}{1+(\frac{i}{10\sqrt 3})^2}$. Prove that 


*

*$x<\frac{\pi}6<y$

*$\frac{x+y}2<\frac{\pi}6$
My try: I have shown that $x,y$ are respectively lower and upper Riemann sum of the function $$I=\sqrt 3 \int_0^1\frac{dx}{3+x^2}=\frac{\pi}6.$$ 
So we are done with part a). How to deal with part b). Here $$x+y=\frac34\frac{1}{10\sqrt 3}+2\sum_{i=1}^{9}\frac{1}{10\sqrt 3} \frac{1}{1+(\frac{i}{10\sqrt 3})^2}+\frac{1}{10\sqrt 3}$$

Write $f(x) = \dfrac{1}{3+x^2}.$  Then part (b) actually follows from the fact
$$0 < x < 1 \implies f''(x) = \frac{6(x^2-1)}{(3+x^2)^3} < 0$$
Graphically, this is because $\frac{x+y}{2}$ can be reinterpreted as the "trapezoid rule" for the integral $\int_0^1 f(x)\,dx.$
