Lower bound on $\int_0^{\pi/4}\sin(\tan x)dx$
Let$$I=\int_0^{\pi/4}\sin(\tan x)dx$$ Show that the following lower bound holds: $$I\ge\frac{\pi^2}{32}$$
Right off the bat, I noticed that $\frac{\pi^2}{32}=\int_0^{\pi/4}xdx$. So the question begs the usage of:
$$\sin x\le x \le \tan x$$
However, trying to apply this leads to a chaotic inequality change:
$$I\ge \int_0^{\pi/4}\sin(x)dx \color{red}{\le}\int_0^{\pi/4}xdx$$
which is inconclusive.
Using your idea, it suffices to show that $\sin(\tan(x)) \ge x$ on $[0, \pi / 4]$. See this question.