Proving $a^ab^b + a^bb^a \le 1$, given $a + b = 1$ Given $a + b = 1$, Prove that $a^ab^b + a^bb^a \le 1$;
$a$ and $b$ are positive real numbers.

Equation $(2)$ is the same as Don Anselmo's proof, but the rest is different.
Bernoulli's Inequality says that for $0\le x\le1$
$$
\left(1+\frac{a-b}b\right)^x\le1+x\frac{a-b}b\tag{1}
$$
multiply by $b$
$$
a^xb^{1-x}\le xa+(1-x)b\tag{2}
$$
substitute $x\mapsto1-x$ in $(2)$
$$
a^{1-x}b^x\le (1-x)a+xb\tag{3}
$$
add $(2)$ and $(3)$
$$
a^xb^{1-x}+a^{1-x}b^x\le a+b\tag{4}
$$
Since $a+b=1$, set $x=a$ and $1-x=b$ in $(4)$
$$
a^ab^b+a^bb^a\le1\tag{5}
$$
