Discuss the inequality different methods inequality $\sum_{k=1}^{n}k^2y_{k}\ge (n+1)\sum_{k=1}^{n}ky_{k}$ 
Nice Problem:   Let $n\ge 3$,and $y_{1},y_{2},\cdots,y_{n}$ be real numbers, and such that $$2y_{k+1}\le y_{k}+y_{k+2}.1<k\le n-2$$
  and $\displaystyle\sum_{k=1}^{n}y_{k}=0$. Show that 
  $$\sum_{k=1}^{n}k^2y_{k}\ge (n+1)\sum_{k=1}^{n}ky_{k}.$$

I have understand  this following nice Proof from AMM11866:
Note \begin{align*}
&0\le\sum_{k=1}^{n-2}\binom{k+1}{2}\binom{n-k}{2}(y_{k+2}-2y_{k+1}+y_{k})\\
&=\sum_{k=1}^{n}\left(\binom{k+1}{2}\binom{n-k}{2}-2\binom{k}{2}\binom{n-k+1}{2}+\binom{k-1}{2}\binom{n-k+2}{2}\right)y_{k}\\
&=\sum_{k=1}^{n}\left(3k^2-3(n+1)k+\binom{n+2}{2}\right)y_{k}\\
&=3\left(\sum_{k=1}^{n}k^2y_{k}-(n+1)\sum_{k=1}^{n}ky_{k}\right)
\end{align*}
Question 1: I want to  know this proof with the coefficient $\binom{k+1}{2}\binom{n-k}{2}$. How to get it?
Question 2: This problem have other methods? (without integral)

The inequality to prove is obviously equivalent to $\sum_{k=1}^{n} k(n-k+1)y_k \le 0$.
Question 1: Idea in the given proof is to use summation by parts to transform the above sum in $y_k$ into a sum in $\Delta_k^{2} \;y_k = y_{k+2}-2 y_{k+1} + y_k$ where the terms are known to be positive by convexity of the sequence. The chosen coefficients "happen" to satisfy the $2^{nd}$ order finite difference equality $$\nabla_{k}^2 \;\binom{k+1}{2} \binom{n-k}{2} = \frac{1}{2}\Big(-6 k(n-k+1) + (n+1)(n+2)\Big)$$
which "conveniently" leads to the sought inequality. Working backwards to actually derive the coefficients from $k(n-k+1)$ is trickier, and requires the appropriate choices of constants in the indefinite sums $\nabla_k^{-1}$, based on the boundary conditions, particular expressions, and general flair.
Question 2: The following will use Chebyshev's sum inequality to prove the stronger statement:

For $n \ge 3$ and $y_k \mid k=1,\cdots,n$ a convex sequence i.e. $y_k \le \frac{1}{2}(y_{k-1}+y_{k+1})$ for $1 \lt k \lt n$:
$$\sum_{k=1}^{n} k(n-k+1)y_k \le \frac{(n+1)(n+2)}{6} \sum_{k=1}^{n} y_k$$

In brief: note first that the coefficients $k(n-k+1)$ are increasing with $k$ for $2k \le n$ $\iff k \le \lfloor \frac{n+1}{2}\rfloor$, and symmetric WRT $k \mapsto n-k+1$. Then group the symmetric terms in $\sum_{k=1}^{n} k(n-k+1)y_k \le 0$ and reduce it to a sum of products of what turns out ot be inversely monotonic sequences, where Chebyshev's inequality applies.
In more detail: $y_k$ is a convex sequence, and it follows from the premise $y_k\le \frac{1}{2}(y_{k-1}+y_{k+1})$ that the finite differences $y_{k+1}-y_{k} \ge y_{k}-y_{k-1}$ form an increasing sequence.
Let $S = \sum_{k=1}^{n} y_k$ and $z_k=y_k+y_{n-k+1}$. Then:


*

*$\quad z_k = z_{n-k+1}$

*$\quad \sum_{k=1}^{n}z_k = 2 \sum_{k=1}^{n}y_k = 2 S$

*$\quad z_{k+1}-z_k = (y_{k+1}-y_k)-(y_{n-k+1}-y_{n-k}) \le 0$ $\iff k \le n-k$ $\iff 2k \le n$ $\iff k \le \lfloor \frac{n+1}{2} \rfloor$ so the sequence $z_k$ is decreasing for $k=1, 2, \;\cdots\;, \lfloor \frac{n+1}{2} \rfloor$
Case 1: Assume $n=2m$ even (so that $\lfloor \frac{n+1}{2} \rfloor = m$) then $\sum_{k=1}^{m}z_k = \sum_{k=1}^{n}y_k = S$ and, by Chebyshev's inequality:
$$
\begin{align}
\sum_{k=1}^{n} k(n-k+1)y_k & = \sum_{k=1}^{m} k(2 m-k+1)z_k \\
 & \le \frac{1}{m}\left(\sum_{k=1}^{m} k(2m-k+1)\right)\left(\sum_{k=1}^{m} z_k\right) \\
 & = \frac{1}{m} \frac{m(m+1)(2m + 1)}{3}  S \\
 & = \frac{(n+1)(n + 2)}{6}  S
\end{align}
$$
Case 2: Assume $n=2m+1$ odd (so that $\lfloor \frac{n+1}{2} \rfloor = m+1$ and $z_{m+1}=2 y_{m+1}$) then:


*

*$\quad \sum_{k=1}^{m}z_k + y_{m+1} = S\;\;$ so $\;\;\sum_{k=1}^{m}z_k = S - y_{m+1}$

*$\quad y_{m+1} \le \frac{S}{2m+1}\;\;$ by Jensen's inequality for abscissae $1,2,\cdots,2m+1$
Again by Chebyshev's inequality:
$$
\begin{align}
\sum_{k=1}^{n} k(n-k+1)y_k & = \sum_{k=1}^{m} k(2 m-k+2)z_k + (m+1)^2 y_{m+1} \\
 & \le \frac{1}{m}\left(\sum_{k=1}^{m} k(2m-k+2)\right)\left(\sum_{k=1}^{m} z_k\right) + (m+1)^2 y_{m+1} \\
 & = \frac{1}{m}\frac{m(m+1)(4m+5)}{6}(S - y_{m+1}) + (m+1)^2 y_{m+1} \\
 & = \frac{m+1}{6}\Big( (4m+5)S + (2m+1)y_{m+1}\Big) \\
 & \le \frac{m+1}{6}\Big( (4m+5)S + S\Big) \\
 & = \frac{(m+1)(2m+3)}{3} S \\
 & = \frac{(n+1)(n+2)}{6} S 
\end{align}
$$
This concludes the proof, and the case $S=0$ gives AMM 11886.
