<s>[INST] Cyclic inequality $\sum_{\text{cyc}}a^ab^bc^c\le1$ how can I prove that for all positive reals $a,b,c$ with $a+b+c=1$ we have 
$$a^ab^bc^c+a^bb^cc^a+a^cb^ac^b\le 1$$
It tried using weighted am.gm. on each term and got $$a^2+b^2+c^2+2ab+2bc+2ac\le 1$$ but now I don‘t know how to proceed.
 [/INST] By weighted AM-GM we can wtite
$$(a^a~b^c~c^c)^{(a+b+c)^{-1}} \le \frac{a.a+b.b+c.c}{a+b+c}~~~(1)$$
and
$$(a^b~b^c~c^a)^{(a+b+c)^{-1}} \le \frac{a.b+b.c+c.a}{a+b+c}~~~(2)$$
and 
$$(a^c~b^a~c^b)^{(a+b+c)^{-1}} \le \frac{a.c+b.a+c.b}{a+b+c}~~~(3)$$
Adding all these equations and using $a+b+c=1$. we get
$$a^ab^bc^c+a^bb^cc^a+a^cb^ac^b \le (a^2+b^2+c^2+2(ab+bc+ca))=(a+b+c)^2=1.$$
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