<s>[INST] 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!
$$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$
 [/INST] Just using calculus.
If you use Stirling approximation, that is to say $$n! \sim \sqrt{2\pi n}\left(\frac n e\right)^n$$ the expression $$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$ becomes $$\frac{e^{a+b}}{2 \pi  \sqrt{a} \sqrt{b}}\le \frac{e^{a+b}}{\sqrt{2 \pi } \sqrt{a+b}}$$ which becomes $$1 \le \frac{\sqrt{2 \pi } \sqrt{a} \sqrt{b}}{\sqrt{a+b}}=\frac{\sqrt{2 \pi a b} }{\sqrt{a+b}}\implies a+b \leq 2\pi a b$$ which looks simpler to analyze.
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