# 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 exercice and a solution, i did my attempt and it was look right , but after some time i remembred 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 verifie 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 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 at 3:30
• I have no idea what the writer was thinking. It might be an oversight. Jul 1 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.