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We are asked to prove or disprove that this is correct $A_1 \cdot B_1 + A_2 \cdot B_2 > A_1 \cdot B_2 + A_2 \cdot B_1$ for $0 < A_1 < A_2$ and $0 < B_1 < B_2$.

I'm not very experienced with writing such proofs, so I'm not sure how should I start, I'm more asking about hints how to write the proof, not really about full proof, however any help would be helpful.

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    $\begingroup$ it's equivalent to $A_2(B_2-B_1)>A_1(B_2-B_1)$ $\endgroup$ – J. W. Tanner May 14 at 20:16
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Note that $$ 0 < (A_2-A_1)(B_2-B_1) $$ and expand the RHS

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Begin with these steps

  1. Move all members to the left side
  2. Make two pairs with common multiplier
  3. Factor out that common multiplier
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A good way is to bring everything on one side and then show that the resulting term is greater than zero. In this case we get $$A_1 \cdot B_1 + A_2 \cdot B_2- A_1 \cdot B_2 - A_2 \cdot B_1$$ We can bring the terms together as follows: $$A_1 \cdot(B_1-B_2) + A_2\cdot (B_2- B_1)$$ and finally this is equal to $$(A_1-A_2)(B_1-B_2)$$ which is greater than zero due to the assumptions ($-\cdot - =+$)

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