How to prove $(n+2)(n+1)k > (n-k+3)(n-k+2)(n-k+1)$ for $k\le n$ and $3k>n+1$

Question

How to prove the following inequality:

$$(n+2)(n+1)k > (n-k+3)(n-k+2)(n-k+1)$$ where $k\le n$ and $3k>n+1$ for integers $n,k$.

Answer 1

Since $3k-1>k$, we obtain $k\geq1$.

In another hand, $k\leq n\leq3k-2$.

Let $f(n)=(n-k+3)(n-k+2)(n-k+1)-k(n+1)(n+2)$.

$f(3k-2)=-k(k-1)(k-2)\leq0$

and $f(k)=-k^3-3k^2-2k+6\leq0$.

$f'(x)=3x^2-(8k-12)x+3k^2-15k+11$,

which says that our inequality is proven for $3k^2-15k+11\geq0$

because there is also $x_1<0$ for which $f(x_1)=0$. (Draw a cubic parabola $y=f(x)$).

Thus, it remains to prove our inequality for $3k^2-15k+11<0$ or $k\in\{1,2,3,4\}$,

which gives that for $k=1$ or $k=2$ your inequality is wrong and for $k\in\{3,4\}$ it's true.

Done!