Prove that $\binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m$ for $m\in \mathbb{N}\backslash \{0,1\}$ Prove the following inequality for every $m\in \mathbb{N}\backslash \{0,1\}$:
$$ \binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m.$$
By some computational arguments, the inequality seems to be true and in particular the difference between the first term $\binom{13+m}{m}$ and the second one $(m+1)\binom{6+m}{m}$ is very large, so obviously is more that $m$. Hence the use of the induction technique, in order to find a way to give a formal proof, leads to a disaster consequence.
In fact, for $m=2$ the inequality is trivial. Fix $m> 2$ and suppose the the inequality is true for $m$. We want to prove that
$$ \binom{14+m}{m+1}-(m+2)\binom{7+m}{m+1}\geq m+1.$$
Start from the first member and apply Stifiel's formula
$$ \binom{14+m}{m+1}-(m+2)\binom{7+m}{m+1}=\binom{13+m}{m+1}+\binom{13+m}{m}-(m+1+1)\Big[\binom{6+m}{m+1}+\binom{6+m}{m}\Big]=
\binom{13+m}{m}-(m+1)\binom{6+m}{m}+\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{6+m}{m+1}-\binom{6+m}{m}\geq...
$$
Applying the inductive hypothesis we have:
$$...\geq m+\binom{13+m}{m+1}-(m+1)\binom{6+m}{m+1}-\binom{6+m}{m+1}-\binom{6+m}{m},$$
but this is less than $m+1$. Therefore we don't get the desired conclusion.
Which can be an useful stategy to prove this inequality?

Using the identity (well-known, easy to prove)
$$\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}, $$
we have
\begin{align*}
 \binom{13 + m}{m} &= \binom{12 + m}{m} + \binom{12 + m}{m - 1}\\
 &= \binom{11 + m}{m} + \binom{11 + m}{m - 1} + \binom{11 + m}{m - 1} + \binom{11 + m}{m - 2}\\
 &\ge \binom{11 + m}{m} + 2\binom{11 + m}{m - 1}\\
 &= \binom{10 + m}{m} + \binom{10 + m}{m - 1}
 + 2\binom{10 + m}{m - 1} + 2\binom{10 + m}{m - 2}\\
 &\ge \binom{10 + m}{m} + 3\binom{10 + m}{m - 1}.
\end{align*}
After repeating the process, we have
$$\binom{13 + m}{m} \ge \binom{7 + m}{m} + 6\binom{7 + m}{m - 1}.$$
It suffices to prove that
$$\binom{7 + m}{m} + 6\binom{7 + m}{m - 1} - (m + 1)\binom{6 + m}{m} \ge m$$
or (after some simple manipulations)
$$6m(m - 1)\frac{(6 + m)!}{m!\, 8!} \ge m$$
or
$$6(m - 1)\frac{(6 + m)!}{m!\, 8!} \ge 1$$
or
$$6(m - 1)\frac{(6 + m)(5 + m)(4 + m)(3 + m)(2 + m)(1 + m)}{8!} \ge 1$$
which is true since
$$\mathrm{LHS} \ge 6(2 - 1)\frac{(6 + 2)(5 + 2)(4 + 2)(3 + 2)(2 + 2)(1 + 2)}{8!} = 3.$$
We are done.
