https://projecteuclid.org/download/pdf_1/euclid.aos/1176350511
Can You help me understand why there is less-than sign in the proof?
$||x-\lambda\alpha_{1}-(1-\lambda)\alpha_{2}||-||x||<\lambda(||x-\alpha_{1}||-||x||)+(1-\lambda)(||x-\alpha_{2}||-||x||)$
Certainly $||x|| = \lambda||x|| + (1-\lambda)||x||$ so the paper is asserting that $$ ||x-\lambda\alpha_1-(1-\lambda)\alpha_2|| < \lambda||x-\alpha_1|| + (1-\lambda)||x-\alpha_2||\;.\tag1$$ This is equivalent to showing $$ ||\lambda(x-\alpha_1)+(1-\lambda)(x-\alpha_2)|| < ||\lambda(x-\alpha_1)|| + ||(1-\lambda)(x-\alpha_2)||\;.\tag2$$ The triangle inequality* says that (2) holds with $\le$ . The reason why the proof excludes the case of equality is the assumption that $x$ does not lie on the line $\ell$ passing through $\alpha_1$ and $\alpha_2$. If the two sides of (2) were equal, then $\lambda(x-\alpha_1)$ would be a scalar multiple of $(1-\lambda)(x-\alpha_2)$, which would imply $x$ lies on $\ell$.
(*)Triangle inequality: In any euclidean space, $||x+y||\le||x||+||y||$ with equality if and only if the triangle formed by $x$, $y$, and $x + y$ is degenerate, that is, $x = 0$ or $y = 0$, or $x=cy$ for some $c$.