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I am trying to prove the following inequality:

$\int_{\tau}^{B} \int_{b}^{A} \frac{a(a-b)}{4a-b} dadb + \int_{\tau}^{B} \int_{\tau}^{b} \frac{b(b-a)}{a-4b} dadb \geq \int_{\tau}^{B} \int_{\tau}^{A} \frac{(a-b)}{4} da db$

I notice that the terms on the left-hand side are somehow symmetrical, but I am not able to use them to simplify the expression. I appreciate any idea that could help me work further on the proof.

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  • $\begingroup$ Can't you just explicitly perform all the integrals? $\endgroup$
    – Michael Engelhardt
    Commented Sep 8, 2021 at 17:44
  • $\begingroup$ What are the relations between $\tau, A, B$? $\endgroup$
    – fedja
    Commented Sep 12, 2021 at 3:00

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