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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.

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1 Answer 1

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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?

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  • $\begingroup$ I <3 u. Seems so obvious after seeing it, as usual. I even had an inkling of that because of the squares and the 2ab. $\endgroup$
    – bean
    Commented Jan 10, 2015 at 5:44
  • $\begingroup$ @Bean: Yeah, even the best have blind spots... don't worry. $\endgroup$
    – Aryabhata
    Commented Jan 10, 2015 at 5:45

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