$a$ and $b$ be positive real numbers such that $a + b = 1$. Prove that $a^a \cdot b^b+a^b\cdot b^a≤1$ 
Let $a$ and $b$ be positive real numbers such that $a + b = 1$. Prove that
  $$a^a\cdot b^b+a^b \cdot b^a≤1$$


By the Jensen's inequality,
$$
\log[a^ab^b]=a\log a+b\log b\leq\log[a^2+b^2]\implies a^ab^b\leq a^2+b^2.
$$
Similarly,
$$
\log[a^bb^a]=a\log b+b\log a\leq \log[ab+ba]\implies a^bb^a\leq 2ab.
$$
Summing then gives:
$$
a^ab^b+a^bb^a\leq a^2+b^2+2ab=(a+b)^2=1.
$$
