Given $3$ positive reals $a$, $b$ and $c$ such that $a+b+c = 1$, show that $a^a b^b c^c + a^b b^c c^a + a^c b^a c^b \le1$.

Question

Good Day! How are you doing? I was learning about the awesome A.M - G.M. inequality from the Brilliant Wiki. There was a question in the exercises:

Given $3$ positive reals $a$, $b$ and $c$ such that $$a+b+c = 1,$$ show that $$a^a b^b c^c + a^b b^c c^a + a^c b^a c^b \le 1$$.

Unfortunately, I was not able to solve this problem nor do I think that solutions are provided. Also, I was not able to find this question here at MSE.

Here was my thought process:

By the AM-GM inequality, it follows that $a^a b^b c^c + a^b b^c c^a + a^c b^a c^b \ge 3(abc)^\frac {1} {3}$

Also, $1 \ge 3(abc)^\frac {1} {3}$ which yields no help.

I also thought of $3(a^a b^ b c^c)^\frac {1} {3} \le a^a + b^b + c^c$ and similarly to other terms but once again, I got stuck at some messy expressions.

I would really appreciate if you could help me.

Thanks!

Answer 1

By the weighted AM-GM inequality, $a^ab^bc^c \leq a \times a + b \times b + c \times c$ (by using weights $a$, $b$, $c$). Similar expressions hold for the other terms. So the inequality becomes $a^ab^bc^c + a^bb^cc^a + a^cb^ac^b \leq a^2 + b^2 + c^2 + ab + bc + ca + ac + ba + cb = (a+b+c)^2 = 1.$

Answer 2

Try to use weighted AM-GM inequality: for any $x,y,z,p,q,r>0$ with $p+q+r=1$ one has $$x^py^qz^r \le px+qy+rz.$$