((a^n+b^n)/2)^( 1/n)>; =(a+b)/2
Where a and b are nonnegative and n is a positive integer.
get
a^n+b^n>; =(a+b)^n*2^( 1-n)
Let a = x 2 and b = y 2.
x^2n+y^2n>; =(x^2+y^2)^n*2^( 1-n)
X 2+y 2 > = (x+y) 2/2, we get
(x^2+y^2)^n*2^( 1-n)>; =(x+y)^2n*2^( 1-2n)
that is
x^2n+y^2n>; =(x+y)^2n*2^( 1-2n)
It can be seen from x+y= 1.
x^2n+y^2n>; =2^( 1-2n)
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