How do I find integers that solve the following inequalities? Give integers $a,b,c,x,y,z$ such that:

$$|a|+|b|+|c| &lt; |x|+|y|+|z|$$ and $$\sqrt{x^2 + y^2 + z^2} \le \sqrt{a^2+b^2+c^2}.$$

I have no idea when this would ever be true, it seems like it would never be true. Any help is appreciated.
Because of the strict inequality involved and the constraint to integer values only, the problem has no solution. To see this, let the ordered tuples of integers $(x,y,z)$ and $(a,b,c)$ represent points in the integer grid $\mathrm R^3.$ Then we can show that the problem $$\sqrt{x^2+y^2+z^2}\le \sqrt{a^2+b^2+c^2}\le |a|+|b|+|c|&lt; |x|+|y|+|z|$$ has no such solutions, by showing that if the sum of coordinate lengths of one of the points, say $A,$ is less than that of the other, say $B,$ then $A$ is strictly closer to the origin than $B,$ contradicting the given conditions.

Thus, without loss of generality let all coordinates be nonnegative (we're in the first octant), and let $p=a+b+c&lt;x+y+z=q.$ Then since we're dealing with integers, assume again without loss of generality that $p=q-1.$ Then any nonnegative three-partition of $p$ will have at least one coordinate less than those of $q.$ It then follows that $A$ is on a sphere of smaller radius than $B.$
Because of the strict inequality involved and the constraint to integer values only, the problem has no solution. To see this, let the ordered tuples of integers $(x,y,z)$ and $(a,b,c)$ represent points in the integer grid $\mathrm R^3.$ Then we can show that the problem $$\sqrt{x^2+y^2+z^2}\le \sqrt{a^2+b^2+c^2}\le |a|+|b|+|c|&lt; |x|+|y|+|z|$$ has no such solutions, by showing that if the sum of coordinate lengths of one of the points, say $A,$ is less than that of the other, say $B,$ then $A$ is strictly closer to the origin than $B,$ contradicting the given conditions.

Thus, without loss of generality let all coordinates be nonnegative (we're in the first octant), and let $p=a+b+c&lt;x+y+z=q.$ Then since we're dealing with integers, assume again without loss of generality that $p=q-1.$ Then any nonnegative three-partition of $p$ will have at least one coordinate less than those of $q.$ It then follows that $A$ is on a sphere of smaller radius than $B.$