03 Mar '08 16:43>
a < b < c < d
prove:
(a + b + c + d)^2 > 8 (ac + bd)
prove:
(a + b + c + d)^2 > 8 (ac + bd)
Originally posted by serigadoConsider
a < b < c < d
prove:
(a + b + c + d)^2 > 8 (ac + bd)
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)