Originally posted by serigadoConsider
a < b < c < d
prove:
(a + b + c + d)^2 > 8 (ac + bd)
{(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)
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)
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
Originally posted by sb3700{a+b+c+d}^2 > 4(a+d)(b+c)
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)
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)