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Re: Ky Fan inequality
Posted:
Oct 21, 2012 9:59 PM


On Sun, 21 Oct 2012, José Carlos Santos wrote:
> 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 > wellknown 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)
f'(0) = 2a(1  a)  a^2 (1  a)  a(2  a)^2 = 2a  2a^2  a^2 + a^3  4a + 4a^2  a^3 = 2a + a^2
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 wellknown > inequalities)? > > Best regards, > > Jose Carlos Santos >



