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




Re: Ky Fan inequality
Posted:
Oct 23, 2012 8:58 AM


On 23102012 1:33, Ken Pledger wrote:
>> .... This first step is: if 0 < a,b <= 1/2, >> then >> >> sqrt(ab)/sqrt((1  a)(1  b)) <= (a + b)/(2  a + b) >> >> .... Can anyone see >> a shorter way of proving this (again, avoiding the use of wellknown >> inequalities)? .... > > > I assume your (2  a + b) is a typo for (2  a  b).
Sure. Thanks.
> Transform the problem a bit. You want to prove that > > (a + b)/sqrt(ab) >= ((1  a) + (1  b))/sqrt((1  a)(1  b)) > > i.e. sqrt(a/b) + sqrt(b/a) >= > sqrt((1  a)/(1  b)) + sqrt((1  b)/(1  a)) > > i.e. sqrt(a/b)  sqrt((1  a)/(1  b)) >= > sqrt((1  b)/(1  a))  sqrt(b/a). > > If you put each side over the appropriate common denominator, then the > numerators come out the same on both sides. Assuming wolog a >= b, > it's easy to show that the common numerator is positive, so cancel it. > Then the necessary inequality between the denominators is also > elementary.
Cute. Thanks.
Best regards,
Jose Carlos Santos



