You have had some good ideas so far. You tried to see when this was true: $\frac{x^3}{3!} \leq \frac{x^4}{4!}.$
You rearranged this to $4x^3\leq x^4$ but you made an incorrect conclusion when you divided by $x^3$ (if $x<0$ then the inequality sign should flip). Instead, lets divide by $x^2$ to get $4x \leq x^2$ or $x(x-4)\geq 0.$ This is true when $x\leq 0$ or $x\geq 4$ so the desired inequality is true in that range.
For $0< x < 4$ we don't have $\frac{x^3}{3!} \leq \frac{x^4}{4!}$ but lets see if the other terms can save us. To do this, we need to see exactly how large $g(x) = x^3/3! - x^4/4!$ can be in $(0,4).$ We calculate that $g'(x) = -(x-3)x^2/6$ so $g$ increases when $0\leq x\leq 3$, the maximum occurs at $g(3)=9/8$, and then it decreases after that.
This is good, because the $1+x+x^2/2$ terms obviously give at least $1$ from $x=0$, and will give us more as $x$ gets bigger. So we solve $1+x+x^2/2=9/8$ and we take the positive solution which is $\frac{\sqrt{5}-2}{2} \approx 0.118.$ So the inequality is definitely true for $x\geq 0.12$ because $g$ is at most $9/8$ and $1+x+x^2/2$ accounts for that amount in that range.
Remember that $g$ was increasing between $x=0$ to $x=3$, so the largest $g$ can be in the remaining range is $g(0.12) = 873/315000 <1$, which is less than the amount $1+x+x^2/2$ gives us. So the inequality is also true for $0\leq x\leq 0.12$, so overall, for all $x.$
So all in all, the only trouble was for $x$ in $(0,4)$ and the contribution from the other terms was always enough to account for $x^3/3!$ when $x^4/4!$ wasn't enough.