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Let x and y be the postive integers. Show that : $\displaystyle\frac{(x + y)!}{ (x + y)^{(x + y)}} < \frac{x! y!}{ (x^x + y^y)}$

Are there any identities we can use to easily prove this?

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up vote 4 down vote accepted

Edit: The statement in itself is not correct: If $x=y=1$, both sides are $\frac{1}{2}$.

Nevertheless, the statement is correct if $x > 1$ or $y > 1$. I'll first prove it for $x,y$ both $>1$, and then for the case where one of them is 1.

I will first assume that $x > 1$, $y > 1$. First, note that $\binom{x+y}{x} = \frac{(x+y)!}{x!y!}$ , so we can rewrite equivalently prove $\binom{x+y}{x} < \frac{(x+y)^{x+y}}{(x^x+y^y)}$.

Next, by the binomial theorem, $(x+y)^{x+y} = \sum_{i=0}^{x+y} \binom{x+y}{i} x^i y^{x+y-i}$. Clearly, $x^x + y^y \le x^x \cdot y^y$, since $x,y > 0$ and hence, $x^x, y^y \ge 1$. Now, $S := \sum_{i=0}^{x+y} \binom{x+y}{i} x^i y^{x+y-i} \ge \binom{x+y}{0} x^0 y^{x+y-0} + \binom{x+y}{x} x^x y^{x+y-x}$ by considering the $0$-th and $x$-th term of the sum (all terms are $\ge 0$). Simplifying this, we get $S \ge 1 \cdot 1 \cdot y^{x+y} + \binom{x+y}{x} x^x y^y > \binom{x+y}{x} x^x y^y \ge \binom{x+y}{x} (x^x + y^y)$. Here, we used that $y^{x+y} \ge 1$.

Dividing by $x^x+y^y$ gives $$ \frac{S}{x^x+y^y} = \frac{(x+y)^{x+y}}{x^x+y^y} > \binom{x+y}{x}. $$ As mentioned above, this proves the claim.

Now assume that $x = 1$ or $y = 1$. I'll do it quickly for $x=1$, the $y=1$ case is symmetric.

We have to prove $$ \frac{(1+y)y!}{(1+y)(1+y)^y} \frac{(1+y)!}{(1+y)^{1+y}} < \frac{1!y!}{1^1+y^y} = \frac{y!}{y^y+1}.$$ Dividing out common factors, this boils down to $$\frac{1}{(1+y)^y} < \frac{1}{y^y+1},$$ which is clear by the binomial theorem and $y > 1$.

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