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.
 A: 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
$$ a^a b^b \leq a^2 + b^2 $$
$$ a^b b^a \leq 2ab $$
$$ \therefore, \qquad a^a b^b + a^b b^a \leq a^2 + b^2 + 2ab = (a +b )^2 = 1^2 = 1 $$
A: 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 $(4)$
$$
a^ab^b+a^bb^a\le1\tag{5}
$$
A: 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.
