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If $f:\mathbb{R} \to \mathbb{R}$ is a strictly increasing function, $f(0) =0$, $a>0$ and $b>0$, then $\int^a_0 f+ \int^b_0 f^{−1} ≥ ab$.

I honestly have no idea how to even begin a formal proof using Riemann integrability. Geometrically, it makes sense to me, but I do not know how to go from there. If anyone could share any hints or help, that'd be great.

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    See https://math.stackexchange.com/questions/1132153/proof-of-youngs-inequality – Tom Chen Feb 28 '19 at 06:57
  • "Geometrically, it makes sense to me" How, exactly? Please, tell us more. – Arthur Feb 28 '19 at 06:57
  • Substitute $u=f^{-1}$ in the second integral. What happens if $a=f(b)$? What does this tell you about the general case? – J.G. Feb 28 '19 at 07:00
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    @Arthur As in, when you think of integration as calculating area. The area of the rectangle with base $a$ and length $b$ is less than the added area calculated in the 2 integrals (due to the intervals we are integrating over with $f$ and $f^{-1}$). –  Feb 28 '19 at 07:12
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    Correction to my previous comment: consider $b=f(a)$. – J.G. Feb 28 '19 at 07:36
  • Do we need continuity for this to be true? –  Mar 01 '19 at 22:51

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