Inequality with increasing variables 
If $n,k\ge2$, and $0\le a_0\le a_1\le\cdots$, prove that
\[\left(\frac{1}{k n} \sum_{l=0}^{k n-1} a_{l}\right)^{k} \geq \frac{1}{n} \sum_{i=0}^{n-1} \prod_{j=0}^{k-1} a_{n j+i}.\]

This inequality is an improvement of AM-GM inequality. For $n=3, k=2$, $0\le a_0\le a_1\le\cdots\le a_5$, the inequality is
\[\left(\frac{a_0+a_1+\cdots+a_5}{6}\right)^2\geq \frac{a_0a_3+a_1a_4+a_2a_5}{3}.\]
This is a homogeneous inequality, so WLOG, let $\displaystyle\sum_{l=0}^{kn-1}a_l=kn$. We have to prove
\[\sum_{i=0}^{n-1} \prod_{j=0}^{k-1} a_{n j+i}\le n.\label1\tag1\]
I tried to use Carlson inequality, it became
\[\left(\sum_{i=0}^{n-1} \prod_{j=0}^{k-1} a_{n j+i}\right)^n\le\prod_{i=0}^{k-1}\sum_{j=0}^{n-1}a_{ni+j}^n\overset?\le n^n.\]
However this doesn't make use of the condition $0\le a_0\le a_1\le\cdots$.
From another perspective we could use the adjustment method. I can prove that $\forall~0\le t\le k-1$, the numbers $a_{tn}$, $a_{tn+1}$ $\ldots$ $a_{(t+1)n-1}$ can take at most three different values to reach maximum.
 A: (With the homogenization of $\sum a_i = kn$.)
First, we consider the $ k = 2, n = 2$ case.
Suppose $ a + b + c + d = 4, a \leq b \leq c \leq d$, we want to show that $ ac + bd \leq 2$.
Intuitvely, if $a \leq d$ are fixed, we'd want to increase $b$ and decrease $c$ subject to $ b \leq c $.
Hence, we replace $(a, b, c, d)$ with $ ( a , \frac{b+c}{2}, \frac{b+c}{2}, d)$, which only increases the value.
This becomes $ a \times \frac{b+c}{2} + \frac{b+c}{2} \times d = (a+d)(b+c) / 2$, which by AM-GM, we know it has a maximum value of $2$, hence we are done.
Equality occurs when $ a + d = b+c = 2$ and $ b = c ( = 1)$. EG $(0, 1, 1, 2)$ and $(1, 1, 1, 1)$ are equality cases.
Note: The hint about the equality condition is that it encourages you to realize that $a+d = b+c = 2$, and so we'd want to force that out somehow.Otherwise, one might think of $(a+b)(c+d)$ instead, but that doesn't help us classify the equality cases, nor have I had success exploring that path.

Now we consider the $k = 3, n = 2$ case.
Suppose $ a + b + c + d + e + f = 6, a\leq b \leq c \leq d \leq e \leq f$, we want to show that $ ace + bdf \leq 2$.
If $a, d, e, f$ are fixed, then since $a \leq d, e\leq f$ so $ae \leq df$ and we'd want to increase $b$ and decrease $c$ subject to $ b \leq c$.
Similarly, if $a, b, c, f$ are fixed, since $ac \leq bf$, we'd want to increase $d$ and decrease $e$ subject to $ d \leq e$.
Thus, we replace $(a, b, c, d, e, f)$ with $(a, \frac{b+c}{2}, \frac{b+c}{2},\frac{d+e}{2}, \frac{d+e}{2},f )$, which only increase the value.
Again, the expression becomes $(a+f)(b+c)(d+e) / 4$, which by AM-GM has a maximum value of 2, and we are done.
Equality holds when $ a +f  = b + c = d+e = 2, b = c ( = 1), d = e (= 1)$.

Now we consider the $k = 2, n = 3 $ case.
Suppose $ a +b + c + d + e + f = 6 , a\leq b \leq c \leq d \leq e \leq f,$ we want to show that $ ad + be + cf \leq 3$.
If $ a \leq f$ are fixed, we'd want to decrease d, increase c subject to $ c \leq d$.
Thus we replace $(a, b, c, d, e, f)$ with $(a, b, \frac{c+d}{2}, \frac{c+d}{2}, e, f)$.

*

*If $ b+e \geq c+d$, then we'd want to replace $(b, e)$ with $(\frac{c+d}{2}, b+e - \frac{c+d}{2})$. Then, we're maximizing $(c+d)/2 \times (6 - 3*(c+d)/2)  $, which has a maximum of $3$ wen $ (c+d)/2 = 1$.

*Otherwise, if $b+e < c+d$, then we'd want to replace $(b, e) $ with $( b+e - \frac{c+d}{2}, \frac{c+d}{2} )$. Then, we're likewise maximizing $(c+d)/2 \times (6 - 3*(c+d)/2)$, which has a maximum of $3$ when $(c+d)/2 = 1$.

The equality cases are

*

*$(a, b, 1, 1, 1, f)$ subject to $ a \leq b \leq 1 \leq f$, $a+b+f = 3$.

*$(a, 1, 1, 1, e, f)$ subject to $ a \leq 1 \leq e \leq f$, $a+e+f = 3$.


I leave you to generalize this, in the exact same way as shown above.
Maybe think about the $ k = 3, n = 3$ case, how can we use the above ideas?
Once you've done that, please post a general solution.
A: Just an idea to simplify the problem :
Take your example :
$$\frac{(a_0+a_1+a_2+a_3+a_4+a_5)^2}{36}\geq \frac{a_0a_3+a_1a_4+a_2a_5}{3}$$
Now if we have :
$$\frac{(a_0+a_1+a_2+a_3+a_4+a_5)^2}{3}\geq \left(a_{2}+a_{5}\right)^{2}+\left(a_{1}+a_{4}\right)^{2}+k\left(a_{0}+a_{3}\right)^{2}$$
Or :
$$\frac{(x+y+z)^2}{3}\geq x^2+y^2+kz^2$$
Or :
$$\frac{(u+v+1)^2}{3}\geq u^2+v^2+k$$
Or :
$$\frac{(u+v+1)^2}{3}-u^2-v^2\geq k$$
On the other hand we need to show :
$$\left(a_{2}+a_{5}\right)^{2}+\left(a_{1}+a_{4}\right)^{2}+k\left(a_{0}+a_{3}\right)^{2}\geq 4(a_0a_3+a_1a_4+a_2a_5)$$
Or :
$$k\left(a_{0}+a_{3}\right)^{2}\ge 4a_0a_3$$
Or :
$$k\geq 1-c^2$$
Remains to show :
$$\frac{(u+v+1)^2}{3}-u^2-v^2\geq 1-c^2$$
Wich is simpler and true if  and $\left(u-1\right)^{2}+\left(v-1\right)^{2}\leq 1-c^{2},c\in[0,1]$
Some details :
$$x=a_2+a_5,y=a_1+a_4,z=a_0+a_3,u=\frac{a_{2}+a_{5}}{a_{0}+a_{3}},v=\frac{a_{1}+a_{4}}{a_{0}+a_{3}},c=\frac{a_3-a_0}{a_0+a_3}$$
PS: We don't need to expand.
