<s>[INST] How to prove $(n+2)(n+1)k > (n-k+3)(n-k+2)(n-k+1)$ for $k\le n$ and $3k>n+1$ 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 $n,k$.
 [/INST] 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!
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