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.

First, a result: for any $x,y$ non-negative reals, $\alpha + \beta = 1 $, then

$$ x^{\alpha}y^{\beta} \leq \alpha x + \beta y $$

To show this, put $f(t) = (1 - \beta) + \beta t - t^{\beta} $. Show this function decreases on $[0,1]$ and then replace $t$ with $\frac{y}{x}$. Now, as for your problem

Notice by application of previous result we get your problem

$$ a^a b^b \leq a^2 + b^2 $$

$$ a^b b^a \leq 2ab $$

$$ \therefore, \qquad a^a b^b + a^b b^a \leq a^2 + b^2 + 2ab = (a +b )^2 = 1^2 = 1 $$

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 $(4)$ $$ a^ab^b+a^bb^a\le1\tag{5} $$

From an algebraic point of view, replace "b" by "1 - a" in the lhs of the inequality. You have a nasty function of "a" but it is workable. The derivative cancels at three values of "a", 0.118007, 0.5 , 0.881993. The maximum of the function corresponds to the middle root (this can be checked using the second derivative test for convexity) and, for a=1/2, the maximum value of the function is equal to 1.