Prove that $A_1 \cdot B_1 + A_2 \cdot B_2 > A_1 \cdot B_2 + A_2 \cdot B_1$ for $0 < A_1 < A_2$ and $0 < B_1 < B_2$

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.

• it's equivalent to $A_2(B_2-B_1)>A_1(B_2-B_1)$ May 14 '19 at 20:16

Note that $$0 < (A_2-A_1)(B_2-B_1)$$ and expand the RHS
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 ($$-\cdot - =+$$)