Expert: Sherry Wallin - 4/3/2009

Question
Answear: Hi Grull~

Try this: since 0<a<1 and 0<b<1 we know that ab<1 and ab^(2/3)<ab<a likewise 0<c<1 and 0<d<1 thus bc^(2/3)< bc < b and cd^(2/3)< cd< c so ab^(2/3)+ bc^(2/3)+ cd^(2/3)<ab + bc + cd < a + b + c < a + b + c + d
= 0 < 3/7



Hello, you are right when a+b+c+d=0. But a+b+c+d=1 (and these numbers are positive). Hope you can help me in this case too, thanks.

    

Make sense?

Math Prof

Answer
Hi Grull~
    Sorry it has taken me a bit to get back to you. Let x^(2/3) be x*. Then try to assume that a is the largest so ab^(2/3)+ bc^(2/3)+cd^(2/3) = ab*+bc*+cd* < a(b*+c*+d*)<=a(b+c+d)* < 3/7[by the triangle inequality] and  by our earlier proof that xy^(2/3) < 3/7 with x = a and y = (b+c+d)^(2/3). Of course there are other cases to consider, like if b is the largest etc...

Math Prof