Processing math: 100%

Friday, November 12, 2021

2021/094) Let a and b be positive real numbers such that a+b=1. Prove that a^a+b^b <=1

We are given 

1= a+ b = a^{a+b} + b^{a+b}

So 1- (a^ab^b + a^b b^a)

=  a^{a+b} + b^{a+b} - (a^ab^b + a^b b^a)

= a^a(a^b-b^b) + b^a(b^b-a^b) = (a^a - b^a)(a^b - b^b)

For a > b both the terms are non -ve so we have and if b > a then both terms are -ve and hence above is positive


1- (a^ab^b + a^b b^a) >=0 and hence the result


No comments: