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)
 A: 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.
A: For question 1: $\binom{k+1}{2}$ and $\binom{n-k}{2}$ are just discrete analogue of $x^2/2$ and $(1-x)^2/2$.  
For question 2: I wouldn't say this is an essentially different proof, but it's the way how I proved it before looking at your proof. Hope it helps :)  
Personally I would like to convert the convexity restriction to the monotonicity of first order difference and apply Abel's summation formula twice given the form of inequality we want to prove. However, there is an additional restriction saying that $\sum y_k=0$. Now we remove the restriction by the following way: since $2y_{k+1}\leq y_k+y_{k+2}$ is invariant under translation $y_k\mapsto y_k+\delta$, the inequality to be proved turns out to be
$$\sum_{k=1}^n k^2(y_k+\delta)\ge(n+1)\sum_{k=1}^n k(y_k+\delta)$$
which simplifies to
$$\sum_{k=1}^n k^2y_k\ge(n+1)\sum_{k=1}^n ky_k+\binom{n+2}{3}\delta$$
with restriction $\sum y_k=n\delta$. So we substitute $\delta$ by $\sum y_k/n$. Hence the inequality is equivalent to
$$\sum_{k=1}^n k^2y_k\ge(n+1)\sum_{k=1}^n ky_k+\frac{1}{3}\binom{n+2}{2}\sum_{k=1}^ny_k$$
Now we arrive at

Claim. For $2y_{k+1}\le y_k+y_{k+2}$ we have
  $$\sum_{k=1}^n k^2y_k\ge(n+1)\sum_{k=1}^n ky_k+\frac{1}{3}\binom{n+2}{2}\sum_{k=1}^ny_k$$
  Proof. Write the inequality as 
  $$\sum_{k=1}^n \left(3k^2-3(n+1)k-\binom{n+2}{2}\right)y_k\ge 0$$
  Applying Abel's summation formula twice easily yields the result.

