Weighted AM-GM Inequality

Question

I was tried to learn some tools on inequalities, and I learn the "Weighted AM-GM Inequality" in a small book: in this book there is an exercise and a solution, I did my attempt and it was look right, but after some time I remembered one thing that make it wrong, but the problem is that the solution in the book used the same idea that I did (and I think this idea is wrong), so I need to verify if I am right or not!

Example 1.2.1. Let $a, b, c$ be positive real numbers such that $a + b + c = 3.$ Show that $a^bb^cc^a\leq1 $

Solution. Notice that$:1=\frac{a+b+c}{3} \geq \frac{ab+bc+ac}{a+b+c}\geq \color{red}{\textrm{$(a^b.b^c.c^a)^{\frac{1}{a+b+c}}$}}$.

I have two notes:

-the red expression must be $\color{red}{\textrm{$(b^a.c^b.a^c)^{\frac{1}{a+b+c}}$}}$.

-the both red expressions are wrong because we can't do this unless $a,b,c$ are positive integers .

this is the definition of Weighted AM-GM Inequality in this book :

Answer 1

1. Perhaps it would be more clear if you rewrote it as $$1 = \frac{a+b+c}{3} \geq \frac{ac + ba + cb}{a+b+c}\geq (c^a \cdot a^b \cdot b^c)^{\frac{1}{a+b+c}}?$$ It's the exact same thing as above, this just emphasizes which weights go where by the order.

2. Weighted AM-GM works even when the weights are real numbers, not just integers. To see this, first notice that weighted AM-GM works with fractional weights. This is because you can multiply each fraction by the lowest common denominator of all the fractions without changing either side of the inequality. Since you can approximate any real number arbitrarily close by fractions, weighted AM-GM works for real numbers too.