Ragnarok7
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This is problem 20b from chapter I 4.10 of Apostol's Calculus I.
The geometric mean $$G$$ of $$n$$ positive real numbers $$x_1,\ldots, x_n$$ is defined by the formula $$G=(x_1x_2\ldots x_n)^{1/n}$$.
Let $$p$$ and $$q$$ be integers, $$q<0<p$$. From part (a) deduce that $$M_q<G<M_p$$ when $$x_1,x_2,\ldots, x_n$$ are not all equal.
(Part (a) was to show that $$G\leq M_1$$ and that $$G=M_1$$ only when $$x_1=x_2=\ldots =x_n$$. I.e., the AM-GM inequality.)
Generalized mean - Wikipedia, the free encyclopedia
Wikipedia mentions it is proved using Jensen's inequality, and all the proofs I have found have used this. But since this comes from the opening of Apostol's book, where calculus has not been developed, I'm wondering if there is a non-calculus way of proving this. Does anyone have any ideas?
The geometric mean $$G$$ of $$n$$ positive real numbers $$x_1,\ldots, x_n$$ is defined by the formula $$G=(x_1x_2\ldots x_n)^{1/n}$$.
Let $$p$$ and $$q$$ be integers, $$q<0<p$$. From part (a) deduce that $$M_q<G<M_p$$ when $$x_1,x_2,\ldots, x_n$$ are not all equal.
(Part (a) was to show that $$G\leq M_1$$ and that $$G=M_1$$ only when $$x_1=x_2=\ldots =x_n$$. I.e., the AM-GM inequality.)
Generalized mean - Wikipedia, the free encyclopedia
Wikipedia mentions it is proved using Jensen's inequality, and all the proofs I have found have used this. But since this comes from the opening of Apostol's book, where calculus has not been developed, I'm wondering if there is a non-calculus way of proving this. Does anyone have any ideas?