Let a,b,c,d be four positive real numbers. Show that $$\sum_{cyc} ab \leq \frac{1}{4}\left(\sum_{cyc} a \right)^2$$ and $$\sum_{cyc} abc \leq \frac{1}{16}\left(\sum_{cyc} a \right)^3$$ My textbook said these inequalities can be referred from the AM-GM inequality but didn't say how to derive it. I can find only looser bounds. That is, for each term in the left hand side of the first inequality, by AM-GM inequality, $$ab \leq \left(\frac{a+b}{2}\right)^2 \leq \frac{1}{4} \left(a+b+c+d\right)^2$$ $$\Longrightarrow \sum_{cyc} ab \leq \frac{4}{4} \left(\sum_{cyc} a \right)^2$$ Using the same techniques, we get another loose bound for $\sum_{cyc} abc$. How could I get the bounds in the original inequalities? Please suggest.
PS. It seems like if these inequalities are true, we may get a general relationship for n positive variables as $\sum_{cyc} a_1 ... a_k \leq \frac{1}{n^{k-1}}\left(\sum_{cyc} a_i \right)^k$.
PS2. Sorry for double posting. There was an old thread here: Showing $\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2$ One answer points to Maclaurin's inequality, but I still cannot figure out how to prove it using AM-GM.
The first inequality.
By AM-GM $$\sum_{cyc}ab=(a+c)(b+d)\leq\left(\frac{a+b+c+d}{2}\right)^2=\frac{1}{4}(a+b+c+d)^2.$$ I used that $$\sum_{cyc}ab=ab+bc+cd+da=b(a+c)+d(a+c)=(a+c)(b+d).$$ The second inequality follows from Maclaurin.
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2,$ where $v>0$ and $abc+abd+acd+bcd=4w^3$.
Thus, by Maclaurin $$u\geq v\geq w.$$
About Maclaurin see here: https://en.wikipedia.org/wiki/Maclaurin%27s_inequality
Your last inequality is wrong.
Try $n=5$ and $k=2$.
But for $n=5$ and $k=3$ it's true.