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How can I prove that forall $0 < x < 1 $ : $$ \frac{2\ln x}{2\arcsin x-\pi} < \frac{\sqrt{1-x^2}}{x} $$ with the Cauchy's mean value theorem that requires continuous in a closed interval and differentiable in a open interval, which I don't have since $\ln 0$ is not defined.

Thanks!

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Define $f(u) =\ln u , h(u) =\arcsin u $ then you have $$\frac{2 \ln u }{2\arcsin u -\pi} =\frac{f(1) -f(x)}{g(1) - g(x) }=\frac{f '(s)}{g ' (s)} =\sqrt{\frac{1}{s^2 } -1 } <\sqrt{\frac{1}{x^2 } -1} $$

since $1>s>x .$

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