# Weighted AM-GM Inequality

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 :

• You can fix the first problem by considering a,b,c in a different order! As for the second, weighted AM-GM is true for nonnegative non-integer powers, though the proof is different. Jul 1, 2022 at 3:25
• But why the writer did not generalize the definition to include any positive power?, does i have to stope learning from this book?
– user1069990
Jul 1, 2022 at 3:30
• I have no idea what the writer was thinking. It might be an oversight. Jul 1, 2022 at 3:57

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.