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I am reading this result which I am not able to prove:

$$ \frac{f'(x)}{f'(f^{-1}(f(x)+a))}-1 $$ is negative for all x and, and for all $a\geq1$ if and only if $f$ is convex. $f'$ is the derivative of $f$ wrt $x$ and $f^{-1}$ is the inverse function. I am actually blanking and do not have an idea about how to prove this result. Can anyone help?

Api
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  • Convex functions are not necessarily invertible, e.g. for $f(x) = x^2$ we do not have an inverse function. – gerw Jan 31 '19 at 14:34

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Try drawing a picture, say of $f(x) = x^2$. To make it invertible, you'll need to restrict your attention to either the right half ($x \ge 0$) or the left half $(x \le 0)$. The top and bottom of that fraction can be seen as slopes of two different tangent lines. Which is steeper?

nkm
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  • I think I get the intuition, since the function is convex then the tangent lines have a smaller coefficient as $x$ increases, therefore their ratio is always smaller than 1. Thank you very much for your help! – Api Jan 30 '19 at 19:38
  • the 'only if' and the technicalities around $a \geq 1$ are still nice challenges – LinAlg Jan 30 '19 at 19:40