<s>[INST] Showing $\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2$ 
$a,b,c,d>0$ with $a+b+c+d=4$ then $\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2$

How can I apply AM-GM here, is there a generalization, say we have $n$ variables summing up to $n$ then,
$\sum\limits_{cyc}a_1\cdot a_2\dots a_k\le\frac1{n^{k-1}}\left(\sum\limits_{cyc}a\right)^k$
because the following inequality should be also true
$\sum\limits_{cyc}abc\le\frac1{16}\left(\sum\limits_{cyc}a\right)^3$ 
 [/INST] Do we need that $a+b+c+d=4$ beacuse if we full expand we get that 
$$a^2+b^2+c^2+d^2+2ac+2bd\geq 2ab+2bc+2cd+2da=2(a+c)(b+d)$$
or
$$(a+c)^2+(b+d)^2\geq 2(a+c)(b+d)$$
Which is true always, or maybe I did a mistake?
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