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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 :

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    $\begingroup$ 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. $\endgroup$ Jul 1 at 3:25
  • $\begingroup$ But why the writer did not generalize the definition to include any positive power?, does i have to stope learning from this book? $\endgroup$
    – user1069990
    Jul 1 at 3:30
  • $\begingroup$ I have no idea what the writer was thinking. It might be an oversight. $\endgroup$ Jul 1 at 3:57

1 Answer 1

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  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.

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  • $\begingroup$ But why did the writer not generalize the definition to include any positive power?, does i have to stope learning from this book? $\endgroup$
    – user1069990
    Jul 1 at 3:30

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