Showing $\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2$

Question

$a,b,c,d>0$ 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

$\sum\limits_{cyc}abc\le\frac1{16}\left(\sum\limits_{cyc}a\right)^3$

Answer 1

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?

Answer 2

By AM-GM $\sum\limits_{cyc}ab=ab+bc+cd+da=(a+c)(b+d)\leq\left(\frac{a+c+b+d}{2}\right)^2=\frac{(a+b+c+d)^2}{4}$.

Your generalization is wrong, but for $k=n-1$ it's true and follows from Maclaurin: https://en.wikipedia.org/wiki/Maclaurin%27s_inequality