Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$ Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$
I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.
$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$ 
How to choose function $f$ and $h.$

You want  to show $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}dx <\frac{\pi}{4}$$
Note that   for $0<x<1$ we have $$ 1+x^2 < 1+x+x^2 <3 $$ Thus $$ \frac {1}{3} <  \frac {1}{1+x+x^2}< \frac {1}{1+x^2}$$
Upon integration we get $$\frac{1}{3}=\int^{1}_0\frac{1}{3}\,dx<\int^1_0\frac{1}{1+x+x^2}\,dx<\int^1_0\frac{1}{1+x^2}\,dx=\arctan(x)\Big|^1_0=\frac{\pi}{4}$$
