<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] It is easy to check that if $a,b$ are two positive real numbers, then $\frac{a^a b^b}{(a+b)^{a+b}}$ is the maximum value of the function $f(x)=x^a(1-x)^b$ over the interval $[0,1]$, attained at $x=\frac{a}{a+b}$. By Euler's beta function we have:
$$ \int_{0}^{1}f(x)\,dx = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}=\frac{1}{a+b+1}\cdot\frac{a! b!}{(a+b)!}\tag{1}$$
Over the interval $\left(0,\frac{a}{a+b}\right)$ the function $(1-x)^b$ is bounded by one, as well as over the interval $\left(\frac{a}{a+b},1\right)$ the function $x^a$ is bounded by one. That leads to:
$$ \int_{0}^{1}f(x)\,dx \leq \int_{0}^{\frac{a}{a+b}}x^a\,dx + \int_{\frac{a}{a+b}}^{1}(1-x)^b\,dx \\=\frac{1}{a+1}\left(\frac{a}{a+b}\right)^{a+1}+\frac{1}{b+1}\left(\frac{b}{a+b}\right)^{b+1}\tag{2} $$
On the other hand, over the interval $\left(0,\frac{a}{a+b}\right)$ the function $(1-x)^b$ is lower-bounded by $\left(\frac{b}{a+b}\right)^b$, as well as over the interval $\left(\frac{a}{a+b},1\right)$ the function $x^a$ is lower-bounded by $\left(\frac{a}{a+b}\right)^a$. That leads to:
$$ \int_{0}^{1}f(x)\,dx \geq \frac{1}{a+1}\left(\frac{a}{a+b}\right)^{a+1}\left(\frac{b}{a+b}\right)^b+\frac{1}{b+1}\left(\frac{b}{a+b}\right)^{b+1}\left(\frac{a}{a+b}\right)^a\tag{3}$$
that is equivalent to:
$$ \int_{0}^{1}f(x)\,dx \geq \frac{a^a b^b}{(a+b)^{a+b}}\left(\frac{\frac{a}{a+1}+\frac{b}{b+1}}{a+b}\right)\tag{4}$$
so:
$$ \frac{a!b!}{(a+b)!}\geq \frac{a^a b^b}{(a+b)^{a+b}}\left(\frac{\frac{a}{a+1}+\frac{b}{b+1}}{a+b}\right)(a+b+1)\tag{5} $$
and:

$$ \frac{a^a b^b}{a! b!}\leq \frac{(a+b)^{a+b}}{(a+b)!}\cdot\frac{a+b}{(a+b+1)\left(\frac{a}{a+1}+\frac{b}{b+1}\right)}\tag{6} $$

is a stronger inequality than the one we wanted to prove, since $(a+b+1)\left(\frac{a}{a+1}+\frac{b}{b+1}\right)\geq (a+b)$.
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