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I am trying to prove the following inequality:

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

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.

I am trying to prove the following inequality:

\int_{\tau}^{B} \int_{e_B}^{A} \frac{e_A(e_A-e_B)}{4e_A-e_B} ,de_A ,de_B + \int_{\tau}^{B} \int_{\tau}^{e_B} \frac{e_B(e_B-e_A)}{e_A-4e_B} ,de_A ,de_B \geq \int_{\tau}^{B} \int_{\tau}^{A} \frac{(e_A-e_B)}{4} ,de_A ,de_B

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.

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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Question on evaluating double integral

I am trying to prove the following inequality:

\int_{\tau}^{B} \int_{e_B}^{A} \frac{e_A(e_A-e_B)}{4e_A-e_B} ,de_A ,de_B + \int_{\tau}^{B} \int_{\tau}^{e_B} \frac{e_B(e_B-e_A)}{e_A-4e_B} ,de_A ,de_B \geq \int_{\tau}^{B} \int_{\tau}^{A} \frac{(e_A-e_B)}{4} ,de_A ,de_B

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.