Maths Question-Help appreciated.Tks?

Question

Suppose that n is a positive integer, and a, b are positive real numbers with a+b=2.
Find the smallest possible value of 1/(1+a^n) + 1/(1+b^n).

Answer 1

Beautiful problem

1/(1+a^n) + 1/(1+b^n)

It is true that as n approaches infinity the number will USUALLY get smaller, but here there is a limitation.

When given such a problem, if "a" is a larger number then b, the first fraction will go to 0 as n approaches infinity and the other one will simplify just go to 1. If b is larger, the second fraction goes to 0 and the first fraction goes to 1.

1 is the actual smallest possible value.

Answer 2

Does this question have to do with limits? Have you studied limits yet? I'll try to explain this easily:

If n is a positive integer, the greater n is, the greater the denominator becomes. The greater the denominator becomes, the smaller the fraction becomes.... So... the smallest possible value would come from "n" being infinity.

If n approaches infinity, the equation will be 1/(infinity) + 1/(infinity). The further "infinity" goes, the bigger the denominator will get, and thus the closer the fraction will get to zero.

So I believe the answer is zero.

better yet:

lim 1/(1+a^n) + 1/(1+b^n) = 0
n--> infinity

Answer 3

English

Answer 4

DL is right. Usually when the expression is symmetric like this, the optimum condition is a=b=1.

You could also do it by calculus:
b = 2-a
f(a) = 1/(1+a^n) + 1/[1 + (2-a)^n]
f'(a) = -n*a^(n-1) / (1+a^n)^2 + n*(2-a)^(n-1) / [1 + (2-a)^n]^2
= 0
By inspection you see that a =1 satisfies f'(a) = 0.

Answer 5

as n approaches infinity then a^n, b^n approach infinity and

1/(1+a^n) + 1/(1+b^n) approach 0

Answer 6

i dont know this math, but im curious. What is this math called like the unit???