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How to prove, that $$f(x)=\frac{||Ax-b||^2}{1-||x||^2}, ||x||^2<1, x \in \mathbb{R}^n$$ is convex?

UPDATE: I've tried the following approach. The function $$g(x) = log(||Ax-b||^2)$$ is convex (proof is straightforward), the same with function $$h(x) = log(\frac{1}{1-||x||^2}), ||x||^2<1$$ The affine combination of $g(x)$ and $h(x)$ is convex too. $$f'(x)=log(\frac{||Ax-b||^2}{1-||x||^2})$$ And applying increasing function $e^x$ preserves the convexity.

Cat-with-a-pipe
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    What's the domain and codomain of $f$? –  Dec 16 '20 at 17:29
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    https://math.stackexchange.com/questions/3834326/prove-that-ax-b-2-1-xtx-is-convex/3834352#3834352 – Michal Adamaszek Dec 16 '20 at 17:29
  • I found an explanation with cones very obscure. I really spend a lot of time, trying to prove, that epigraph of $f$ is a cone, but i failed – Cat-with-a-pipe Dec 17 '20 at 20:18
  • The idea is to describe the set you want as intersections and linear projections of sets you already know are convex, in this case quadratic cones (and hyperplanes, halfspaces etc.). That is often a very natural strategy, and the presence of 2-norms in $f$ is just asking for it. I am pretty sure that any direct proof of convexity of $f$ will at some point be reproving the convexity of those basic building blocks. – Michal Adamaszek Dec 18 '20 at 08:28
  • I agree that separating the whole problem on blocks looks very natural here. But still, I feel very uncomfortable with cones :( @MichalAdamaszek can I ask you to criticize my approach to the proof (I've added the update) – Cat-with-a-pipe Dec 19 '20 at 16:38
  • g is not convex. – Michal Adamaszek Dec 21 '20 at 15:18
  • Yep, I missed it. Thanks! – Cat-with-a-pipe Dec 21 '20 at 16:56
  • Can you please explain $s > ||x||^2$. How it appeared? – Cat-with-a-pipe Dec 21 '20 at 17:06

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