a<b and c<d mean that 1/a>1/b and 1/c>1/d. Thus 1/ac>1/bd.
b<d means that d-b>0 and b-d<0, but c>a, so a(d-b)+c(b-d)=ad-ab+bc-cd<0.
Thus (ad-ab+bc-cd)/ac<(ad-ab+bc-cd)/bd.
Factoring and rearranging the right side of the inequality, we get [a(d-b)+c(b-d)]/ac<[d(a-c)+b(c-a)]/bd.
Splitting each rational expression and cancelling factors, we get (d-b)/c +(b-d)/a < (a-c)/b + (c-a)/d.
Splitting expressions again: d/c - b/c + b/a - d/a < a/b - c/b + c/d - a/d.
Add b/c + c/b + d/a + a/d to both sides to get d/c + b/a + c/b + a/d < a/b + c/d + b/c + d/a, which is the desired result.
How's that?
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Posted by Mike C
on 2007-11-08 10:39:02 |
Trying again
