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Topic:
Ky Fan inequality
Replies:
5
Last Post:
Oct 25, 2012 5:02 AM




Ky Fan inequality
Posted:
Oct 21, 2012 8:20 AM


Hi all,
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)
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.
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



