How can I prove $(a+b+c)!>a!b!c!$ In fact, I couldn't prove the inequality because I don't know which method is used for this.
The condition for this inequality is  $$(a+b+c)>1$$  
 A: This will help:
$$(a+b)! = a!(a+1)(a+2)...(a+b)<a!(1)(2)...(b)=a!b!.$$
Your turn to prove it for $(a+b+c)!$...
$$(a+b+c)! = a![(a+1)(a+2)...(a+b)][((a+b)+1)((a+b)+2)((a+b)+c)].$$
A: Show that $(x+y)! > x!y!$:
Divide by $x!$ to get $(x+1)(x+2)(x+3)\dots(x+y) > y! = (1)(2)(3)\dots(y)$, which is true for $x > 0$ (compare terms pairwise!)
Let $x = a, y = b+c$. Then we have $(a+b+c)! > a!(b+c)!$
Furthermore we have $(b+c)! > b! c!$, so $(a+b+c)! > a!(b+c)! > a!b!c!$.
A: $$\frac{(a+b+c)!}{a!b!c!} $$
is the number of ways to arrange $a$ (indistinguishable) amber balls, $b$ blue balls and $c$ cyan balls in a line. Your goal is to show that this number is $>1$. This is the case as soon as at least two of $a,b,c$ are positive.
A: $(a+b+c)!$ and $a!b!c!$ are both products of $a+b+c$ terms. Can you see a way of comparing them individually?
A: Hint: 
$$(a+b+c)!= a! \prod_{k=a+1}^{a+b} k! \prod_{j=a+b+1}^{a+b+c}j!, $$
where $\prod_{k=a+1}^{a+b}$ is products of $b$ terms each of which is at least $b$,
hence $\prod_{k=a+1}^{a+b}\geq b^b>b!$...
Similarly for $\prod_{j=a+b+1}^{a+b+c}j!>c!$.
A: Use induction to show that $(a+b)! > a!b!$. You'll need to induce over both a and b. Note that this is for $a \ge 1, b \ge 1$.
A: An adaptation of this answer. If you have $a$ adults, $b$ boys, and $c$ girls, then $(a+b+c)!$ counts the number of ways you can line everyone up, while $a!b!c!$ counts the number of ways you can do so with the additional requirement that all adults come before all children, and all boys before all girls. Then $(a+b+c)!\geq a!b!c!$ always holds, while $(a+b+c)!>a!b!c!$ holds as soon as at least two categories are non-empty, since then there will be some line-up that violates the additional requirement.
