# Inequality

Posers and Puzzles 03 Mar '08 16:43
1. 03 Mar '08 16:43
a < b < c < d

prove:
(a + b + c + d)^2 > 8 (ac + bd)
2. coquette
03 Mar '08 21:35
no
3. 03 Mar '08 23:08
Originally posted by coquette
no
no? it's wrong?
4. 04 Mar '08 16:50
The rearrangement inequality. cbf working it out though
5. wolfgang59
04 Mar '08 20:53
a < b < c < d

prove:
(a + b + c + d)^2 > 8 (ac + bd)
Consider
{(a+b)-(c+d)}^2 > 0

Expand
(a+b)^2 + (c+d)^2 - 2(a+b)(c+d) > 0

* Therefore
(a+b)^2 + (c+d)^2 > 2(a+b)(c+d)

Now
{a+b+c+d}^2 = {(a+b)+(c+d)}^2
= (a+b)^2 + (c+d)^2 + 2(a+b)(c+d)

Now substituting for (a+b)^2 + (c+d)^2 from * gives;

{a+b+c+d}^2 > 4(a+b)(c+d)
expand

{a+b+c+d}^2 > 4(ac + bc + ad + bd)
{a+b+c+d}^2 > 4(ac + bc) + 4(ad + bd)

Let b=a+x (where x is =ve since we know b>a)

Substitute
{a+b+c+d}^2 > 4ac + 4(a+x)c + 4(b-x)d + 4bd

{a+b+c+d}^2 > 8ac + 4xc + 8bd -4xd

{a+b+c+d}^2 > 8(ac + bd) + 4x(c-d)

Therefore
{a+b+c+d}^2 > 8(ac + bd)
6. wolfgang59
04 Mar '08 21:03
Just one flaw with my proof.
Its wrong!
ðŸ˜ž
7. 06 Mar '08 06:271 edit
Originally posted by wolfgang59
{a+b+c+d}^2 > 8(ac + bd) + 4x(c-d)
as d>c, c-d < 0 so next step is flawed
8. 06 Mar '08 06:57
my solution, based on wolfgang:

Consider
{(a+d)-(b+c)}^2 > 0

Expand
(a+d)^2 + (b+c)^2 - 2(a+d)(b+c) > 0

* Therefore
(a+d)^2 + (b+c)^2 > 2(a+d)(b+c)

Now
{a+b+c+d}^2 = {(a+d)+(b+c)}^2
= (a+d)^2 + (b+c)^2 + 2(a+d)(b+c)

Now substituting for (a+d)^2 + (b+c)^2 from * gives;

{a+b+c+d}^2 > 4(a+d)(b+c)
expand

{a+b+c+d}^2 > 4(ab + ac + bc + bd)

Consider - ab + ac - cd + bd
= b(d-a) - c(d-a)
= (b-c)(d-a)
4(ab + ac + bc + bd):
{a+b+c+d}^2 > 4(ab + ac + bc + bd - ab + ac - cd + bd)
{a+b+c+d}^2 > 4(2ac + 2bd)

Hence:
{a+b+c+d}^2 > 8(ac + bd)
9. 06 Mar '08 13:035 edits
Consider - ab + ac - cd + bd
= b(d-a) - c(d-a)
= (b-c)(d-a)

You only forgot to say this quantity is negative because a < b< c< d

4(ab + ac + bc + bd):
{a+b+c+d}^2 > 4(ab + ac + bc + bd - ab + ac - cd + bd)
{a+b+c+d}^2 > 4(2ac + 2bd)

And forgot to mention bc - cd is negative too, so the inequality is maintained
10. 06 Mar '08 14:271 edit
11. 06 Mar '08 14:55
Originally posted by sb3700
my solution, based on wolfgang:

Consider
{(a+d)-(b+c)}^2 > 0

Expand
(a+d)^2 + (b+c)^2 - 2(a+d)(b+c) > 0

* Therefore
(a+d)^2 + (b+c)^2 > 2(a+d)(b+c)

Now
{a+b+c+d}^2 = {(a+d)+(b+c)}^2
= (a+d)^2 + (b+c)^2 + 2(a+d)(b+c)

Now substituting for (a+d)^2 + (b+c)^2 from * gives;

{a+b+c+d}^2 > 4(a+d)(b+c)
expand

{a+b+c+d}^2 > 4(ab ...[text shortened]... c + bc + bd - ab + ac - cd + bd)
{a+b+c+d}^2 > 4(2ac + 2bd)

Hence:
{a+b+c+d}^2 > 8(ac + bd)
{a+b+c+d}^2 > 4(a+d)(b+c)
expand

This far I understand, but it gets confusing from here on. Where does the bc term come from in the next line?

{a+b+c+d}^2 > 4(ab + ac + bc + bd)

Consider - ab + ac - cd + bd
= b(d-a) - c(d-a)
= (b-c)(d-a)
12. 06 Mar '08 16:01
I get it now. Your proof looks good apart from the mix up with bc and cd.