$a$ 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
  $$a^a\cdot b^b+a^b \cdot b^a≤1$$


Hint: Since $a + b = 1$, we can use the Weighted Arithmetic Mean - Geometric Mean inequality to get $$\sqrt[a+b]{a^a b^b} \leq a^2 + b^2.$$ Similarly, write the inequality for the other term to get the desired result.
