2019/015) Let a and b be positive real numbers such that $a+b=1$. Prove that $a^ab^b +a^bb^a <=1$

without loss of generality we can assume $a>=b$
We have
$1= a+ b = a^{a+b} + b^{a+b}$
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

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