If 0 < a < b < c < d then prove that

a/b + b/c + c/d + d/a > b/a + c/b + d/c + a/d.

d>b implies b-d<0, d-b>0

c>a implies abs(b-d)/c<abs(d-b)/a. Thus (d-b)/a + (b-d)/c > 0.

Similarly, we can show that (c-a)/d + (a-c)/b > 0.

Adding the inequalities together and splitting fractions, we get:

d/a - b/a + b/c - d/c + c/d - a/d + a/b - c/b > 0

Adding b/a + d/c + a/d + c/b to both sides provides the desired result.

Posted by Mike C on 2007-11-07 11:13:16