Cyclic inequality $\sum_{\text{cyc}}a^ab^bc^c\le1$

Question

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.

Answer 1

We have $$a^2+b^2+c^2+2ab+2bc+2ac=(a+b+c)^2=1$$ which proves your inequality.

Supposing that $a,b,c\neq 0$ we see that equality in the weighted AM-GM occurs if and only if $a=b=c=\frac13$.

Answer 2

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 $$a^ab^bc^c+a^bb^cc^a+a^cb^ac^b \le (a^2+b^2+c^2+2(ab+bc+ca))=(a+b+c)^2=1.$$