> Please consider the Ky Fan inequality: > http://en.wikipedia.org/wiki/Ky_Fan_inequality > > I tried to prove it directly (that is, without using some other > well-known inequality) by induction and already the first step was > harder than what I expected. This first step is: if 0 < a,b <= 1/2, > then > sqrt(ab)/sqrt((1 - a)(1 - b)) <= (a + b)/(2 - a + b)
Since all quantities are positive, you're wanting to prove (2 - a + b).sqr ab <= (a + b).sqr (1 - a)(1 - b) or ab(2 - a + b)^2 <= (a + b)^2 (1 - a)(1 - b).
Let f(x) = (a + x)^2 (1 - a)(1 - x) - ax(2 - a + x)^2.
f'(x) = 2(a + x)(1 - a)(1 - x) - (a + x)^2 (1 - a) - (a(2 - a + x)^2 + 2ax)
0 <= f'(0) iff 0 < -2 + a iff 2 < a. That's not a good sign for b close to 0.
> What I did was to square both sides and then what I needed to prove > was that > > (a + b)^2/(2 - a - b)^2 - ab/((1 - a)(1 - b)) >= 0.
Hey, you changed a sign of b.
> So, I turned the LHS of this expression into a rational expression, > whose denominator is clearly >= 0 and whose numerator is > > a^2 - a^3 + b^2 - b^3 + a^2b + ab^2 - 2ab. (*) > > Since the original inequality is actually an equality when a = b, this > is also true for (*) and so I divided (*) by a - b, getting > > a - a^2 - b + b^2 = a - b - (a^2 - b^2) = (a - b)(1 - a - b). > > So, (*) = (a - b)^2(1 - a - b), which is clearly >= 0. > > I have the feeling that what I did was too complicated. Can anyone see > a shorter way of proving this (again, avoiding the use of well-known > inequalities)? > > Best regards, > > Jose Carlos Santos >