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If $$a<b<c<d$$ Which is larger without using examples of numbers?

$$x=(a+b)(c+d), y=(a+c)(b+d), z=(a+d)(b+c)$$

I did this exercise, but I have not managed to complete it. The first thing I did was to expand them and then compare ax with z and so on, I know that the correct answer is that z is greater, but I can not assume I have to prove it and my doubt is at the end of taking away the similar terms, since I am assuming that the difference of one side is less than the term that remains for the other.

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In order to solve this you can expand the three expressions, and then compute $z-x$ and $z-y$ which are respectively equal to $(d-b)(c-a)$ and $(d-c)(b-a)$ which are both positive. Therefore $z$ is the largest of all three variables.

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Here is a proof $x \lt y$

$$ x = ac + ad + bc + bd$$

$$ y = ab + ad + cb + cd$$

Let us start with assuming $x < y$

$$ x = ac + ad + bc + bd \lt ab + ad + cb + cd = y$$

Cancelling out like terms:

$$ ac + bd \lt ab + cd $$

And then rearranging:

$$0 \lt cd - ac + ab -bd$$

$$ 0 \lt c(d-a) +b(a-d)$$

For simplicity, let's define g = d-a

Then

$$0 \lt cg - bg$$

$$ 0 \lt g(c-b)$$

Where $g$ is positive and $c-b$ is positive, so indeed

$$ 0 \lt g(c-b)$$

is true and thus, the assumption $x \lt y$ is true.

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Hint: assuming $$x\le y$$ we get by expanding $$ac+bc+ad+bd\le ab+bc+ad+cd$$ then we get $$c(a-d)\le b(a-d)$$ so we get $$(a-d)(c-b)\le 0$$ Can you finish? since $a-d<0$ and $c-b>0$ then our inequality is true.

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  • $\begingroup$ In that part I stayed because by solving them how can I know that ab +cd>ac+bd? $\endgroup$ – Sonia f May 3 '18 at 15:53
  • $\begingroup$ And I think there's an error in what you said instead of a-d it would be a-b $\endgroup$ – Sonia f May 3 '18 at 15:54
  • $\begingroup$ Something went wrong. (a-d) is negative and (c-d) is negative. Which means (a-d)(c-d)>0. Which is a contradiction, which implies the original assumption x<y is false. However, testing some real numbers shows x<y is true. $\endgroup$ – CEP May 3 '18 at 16:01
  • $\begingroup$ @sonia $ a-d $ is correct. In the first inequality cancel out the terms. Then rearrange it by subtracting the right terms. $\endgroup$ – Patrick Abraham May 3 '18 at 16:03
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Let $c-b = k>0$

Then $y=(a+c)(b+d) = $

$(a+b+k)(c + d -k) = $

$(a+b)(c+d) + k[c+d - a-b] -k^2=$

$x + k[d-a + k] - k^2=$

$x + k(d-a) > x$.

So $y > x$

Let $d-c = m>0$

$z = (a+d)(b+c)=$

$(a + c+m)(b+ d - m)=$

$(a+c)(b+d) + m[b+d - a-c] - m^2=$

$y + m[b-a +m] - m^2 =$

$y + m(b-a) > y$.

So $z > y$

So $z > y > x$.

....

Also there is AM-GM

If $j < k,m; k,m < n$ then $nj > km$ so

$(a+b) < (a+c)$ and $(b+d)< (c+d)$ so $(a+b)(c+d) > (a+c)(b+d)$... but ... I don't know. That doesn't have as "hands on" conviction. (Although its really the same thing.

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