The special factorial inequality

The given inequality holds when at least two of $x,y,z$ are $0$ .

If two of them is $0$ , the cases are $1. \;x=y=0$ , $z=1$ to $10$ . Similarly $2. \;y=z=0$ , $x=1$ to $10$ and $3. \;z=x=0$ , $y=1$ to $10$ .

If all of $x=y=z=0$ , then $1$ solution exists.

So, total cases are $3*10+1=\boxed{31}$

The given inequality does not hold when each of $x,y,z$ is greater than $0$ .

Without loss of generality, let $x \geq y \geq z$ . Thus, $z=0$ .

So, we are left with $x! + y! +1 \geq (x+y)!$

Clearly, when each of $x,y$ is greater than $1$ , the inequality does not hold.

So we have two cases to consider:-

• $y = 0$ . Here, we get $x!+2 > x!$ , which is true for all nonnegative integers.Clearly, for $x \in [1,10]$ , we get $(x,0,0)$ as an ordered triple. Considering the permutations for $x,y,z \in [1,10]$ we get $30$ ordered triplets. But, since $(0,0,0)$ satisfies the inequality, we get a total of $31$ ordered triples for $x,y,z \in [0,10]$

• $y = 1$ . Here, we get $x!+2 > (x+1)!$ , which holds when $x$ is less than $2$ . Thus, $x=1$ and $(1,1,0)$ is an ordered triplet. Considering the permuations for $x,y,z \in [0,10]$ , we get $3$ ordered triplets.

Therefore, $34$ ordered triplets satisfy the given inequality.