hi,
Could somebody please give me an idea of how to prove the following inequality:
a/b + b/c + c/d + d/a >= b/a + c/b + d/c + a/d
a,b,c,d being real numbers and 0 < a < b < c < d
thanx
Consider doing it by parts : prove that :
$\displaystyle \frac{a}{b} \geq \frac{a}{d}$, since $\displaystyle a < b < d$
Prove the three other pairs, then add everything together (two or more inequalities of the same sign can be added up). You should prove your inequality just fine
Everything's positive here, so multiply both sides by their common denominator and then you must prove that
$\displaystyle a^2cd+b^2ad+c^2ab+d^2bc\ge b^2cd+c^2ad+d^2ab+a^2bc\Longleftrightarrow$ $\displaystyle a^2c(d-b)+b^2d(a-c)+c^2a(b-d)+d^2b(c-a)\ge 0\Longleftrightarrow$ $\displaystyle (c-a)bd(d-b)+(d-b)ac(a-c)\ge 0$
$\displaystyle \Longleftrightarrow (c-a)(d-b)(bd-ac)\ge 0\Longleftrightarrow$ $\displaystyle bd\ge ac$ , and since this last inequality is (almost trivially) true you can go back and prove the original one...
Tonio