Using binomial Theorem how we can show $\frac{(x+y)!}{x!y!}\leq \frac{(x+y)^{x+y}}{x^xy^y}$?

Question

Using binomial Theorem prove that $$\frac{(x+y)!}{x!y!}\leq \frac{(x+y)^{x+y}}{x^xy^y}.$$

I tried it as follows: It is clear that $x\leq x+y, \forall x,y\in \mathbb{N}$. Thus, by Binomial Theorem, we have \begin{align*}(x+y)^{x+y}&=\displaystyle\sum_{y=0}^{x+y}{x+y\choose y}x^{(x+y)-y}y^{y}\\&=\sum_{y=0}^{x+y}\frac{(x+y)!}{x!y!}x^{x}y^{y}\\&\geq \frac{(x+y)!}{x!y!}x^{x}y^{y}(how?)\end{align*}

I can't show the last inequality, thus is there any one who can give me hint over here, please? Thanks .

Answer 1

Hint: Expand $(x+y)^{x+y}$ and look at the $(y+1)^{th}$ term. Does it equal your LHS?

Answer 2

$$(x+y)^{x+y}=\sum_{k=0}^{x+y}\frac{(x+y)!}{k!(x+y-k)!}x^{x+y-k}y^k\ge\underbrace{\frac{(x+y)!}{y!(x+y-y)!}x^{x+y-y}y^y}_{\text{evaluated at}\,k=y}=\frac{(x+y)!}{x!y!}x^xy^y$$