How to prove that $\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$? As stated in the question. Thank you!
$$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$

Base case: Assume $a=b=0$. Then $\frac{0^00^0}{0!0!}\le\frac{(0+0)^{(0+0)}}{(0+0)!}$ goes to $1 \le 1$, which is true.
Inductive case: Suppose $a=b$. Increase $b$ to equal $a+k$ for positive $k$. Then $\frac{a^a(a+k)^{a+k}}{a!(a+k)!}\le\frac{(a+a+k)^{(a+a+k)}}{(a+a+k)!}$ rearranges to $\frac{(2a+k)!}{a!(a+k)!} \le \frac{(2a+k)^{2a+k}}{a^a(a+k)^{a+k}}$ at which point you can compare each piece: $(2a+k)! \le (2a+k)^{2a+k}$, $a! \le a^a$, and $(a+k)! \le (a+k)^{a+k}$.
The same can be said for setting $a$ equal to $b+k$.
