((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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