Prove/refute property about a constrained sequence of numbers I need to prove or refute a property about a sequence of numbers.
Here is what is given to me:
Sequence ($a_1,a_2,...,a_k,a_{k+1},a_{k+2}$) containing $k+2$ numbers. Every number $0 &lt; a_i \leq M, i=1,...,k+2$ for the same given constant $M$. Moreover, $\sum_{i=1}^{k+2} a_i = N$, for another given constant $N &gt; 0$. The constants $N$ and $M$ are related by $kM \geq N, k &gt; 1$.
Then, I need to prove or refute the following property: Can all consecutive pairs of numbers $a_i, a_{i+1}, i=1,...,k+1$ be defined such that $a_i + a_{i+1} &gt; M$ ? Or does it lead to a violation of one of the given constraints?
I tried searching for something similar but in truth, I barely know what to search for. My tentative proofs do not really go anywhere meaningful, so I had to resort to more experienced math guys to help me with this. It has been a long time since I had to prove some property like this.
Assuming that given $k$, $N$ is selected such that
$$\sum_{i=1}^{k+2} a_i = N,$$
I claim it is possible $ \iff k\ge 3$
First of all we see the cases $k=1,2$ to check it is impossible:
(i) if $k=1$ then
$$M&lt; a_1+a_2 &lt; N \le kM = M;$$
(ii) if $k=2$ then we only have $(a_1,a_2,a_3,a_4)$, and
$$a_1+a_2&gt;M \mbox{ and } a_3+a_4 &gt;M ,$$
so
$$N = \sum_{i=1}^{k+2} a_i &gt;M + M =2M = kM .$$
Now let $k\ge 3$: let $\epsilon &gt; 0$ (arbitrarily small) and such that $\epsilon &lt; M$, then I pick:
$$a_i= M \mbox{ if } i \mbox{ is even; } a_i = M -\epsilon \mbox{ if } i \mbox{ is odd.}$$
Now we obseve that $a_i + a_{i+1} = 2M - \epsilon &gt; M$ anf that $0 &lt; a_i \le M$, so they are good, then:
(i) if $k$ is even
$$\sum_{i=1}^{k+2} a_i = N = (2M -\epsilon)\Big(\frac{k}{2} + 1\Big) = (k+2)M - \epsilon\Big(\frac{k}{2}+1\Big)$$
which is possible if I pick $\epsilon$ such that
$$kM \ge (k+2)M - \epsilon\Big(\frac{k}{2}+1\Big)$$
which implies that
$$\epsilon \ge \frac{2M}{\frac{k}{2}+1} .$$
Also we have that $M&gt; \epsilon$, so we can pick such a $\epsilon$ only if
$$M&gt; \frac{2M}{\frac{k}{2}+1},$$
which means that $k &gt; 2$.
(ii) if $k$ is odd
$$\sum_{i=1}^{k+2} a_i = N = (2M-\epsilon)\Big(\frac{k+1}{2}\Big) + a_{k+2} = (k+1)M - \epsilon \Big(\frac{k+1}{2}\Big) + a_{k+2} =$$
$$= (k+2)M - \epsilon \Big(\frac{k+1}{2} +1 \Big)$$
and here too we see it is possible if $\epsilon$ is such that
$$\epsilon \ge \frac{2M}{\frac{k+1}{2}+1} .$$
This implies that it is tue if
$$M &gt; \frac{2M}{\frac{k+1}{2}+1},$$
which means if $k&gt;1$.
This concludes my claim.
If $a_i+a_{i+1}&gt;M$ for all applicable $i$, then
$$ N=\sum_{i=1}^{k+2}a_i=\frac{a_1}2+\sum_{i=1}^{k+1}\frac{a_i+a_{i+1}}2+\frac{a_{k+2}}2&gt;\frac{k+1}2M$$
and in fact
$$ N=\sum_{i=1}^{k+2}a_i=\sum_{j=1}^{\frac{k+2}2}(a_{2j-1}+a_{2j})&gt;\frac{k+2}2M\qquad\text{if $k$ is even}.$$
Hence we certainly need the additional condition that
$$\tag1 \left\lceil\frac{k+1}2\right\rceil M&lt;N.$$
In particular, $kM\ge N$ contradicts $(1)$ when $k=1$ or $k=2$. Hence we incidentally also need
$$\tag2 k\ge3 $$
(but as said, $(1)$ implies $(2)$ in the given context).
Finally, we also need
$$ \tag3 M&gt;0$$
to allow for $0&lt;a_i\le M$ in the first place.

On the other hand, assume we have $k\in\Bbb N$, $N,M\in \Bbb R$ such that $(1)$ and $(3)$ and $kM\ge N$. Then we can let
$$a_i=\frac N{k+2}. $$
By $(1)$ and $(3)$ and $N\le kM$, this makes
$$ 0&lt;a_i&lt;M.$$
We clearly have
$$ \sum a_i = (k+2)\cdot \frac N{k+2}=N$$
and by $(1)$,
$$ a_i+a_{i+1}=\frac{2N}{k+2}\ge\frac N{\left\lceil \frac{k+1}2\right\rceil}&gt;M,$$
as desired.