How can i prove that this function is in the given bounds? So, I have been struggling to show the following:
Let $a \in [0,1)$, $b \in \left[0,\frac{1-a^2}{1+a^2}\right)$, and $f(a,b)=\sqrt{a^2+b^2-a^2b^2 + 2ab\sqrt{1-b^2}}$.
Prove that $a \leq f(a,b) < 1$
The $a \leq f()$ is pretty trivial, but the other inequality escapes me.
I have verified that $f(a,0)=a$ (obvious) and $f\left(a,\frac{1-a^2}{1+a^2}\right)=1$, and $f$ appears to be monotone increasing in $b$ for each $a$, but this just led to another bound that was difficult to show.

Notice that:
$$ \left(a\sqrt{1-b^2} + b\right)^2 = a^2(1-b^2) + b^2 + 2ab\sqrt{1-b^2} = a^2 + b^2 - a^2b^2 + 2ab\sqrt{1-b^2}$$
Thus your expression is $a\sqrt{1-b^2} + b$
Can you take it from here?
