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So, I'm reading the literature to find different proofs of the AM-GM inequality, the following proof quite hit me, and I don't seem to understand at all. The proof is as follows:

For any positive numbers: $a_1,a_2,...a_n$. We have: $$ \dfrac{a_1+a_2+..+a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 .... \cdot a_n} $$

Replacing $a_k$ with $a_k \sqrt[n]{a_1 \cdot a_2 .... \cdot a_n}$ for $1\leq k \leq n$ we have $a_1 \cdot a_2 ... \cdot a_n = 1$, so that is enough to prove that $ {a_1+a_2+..+a_n} \geq n$.

My question is: why $a_1 \cdot a_2 ... \cdot a_n = 1$ ? And, in general, for any sequence of positive numbers $ a_k \neq a_k \sqrt[n]{a_1 \cdot a_2 .... \cdot a_n}$?

I would really appreciate some light in this matter.

P.s I'm annexing a photo of the section.

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I think they mean replacing $a_k$ by $\dfrac{a_k}{\sqrt[n]{a_1a_2\cdots a_n}}$. Another way to see that is to divide both side of the original AM-GM inequality by $\sqrt[n]{a_1a_2\cdots a_n}$ to get an equivalent inequality $$\frac1n\left(\frac{a_1}{\sqrt[n]{a_1a_2\cdots a_n}}+\frac{a_2}{\sqrt[n]{a_1a_2\cdots a_n}}+\cdots+\frac{a_n}{\sqrt[n]{a_1a_2\cdots a_n}}\right)\ge 1,$$ which is AM-GM for $\dfrac{a_k}{\sqrt[n]{a_1a_2\cdots a_n}}$.

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  • $\begingroup$ Oh wow nice spot, do you happened to know why $a_1 < 1 < a_2 $ ? $\endgroup$ – David Cardozo Oct 28 '14 at 5:32
  • $\begingroup$ Since $a_1\cdots a_n=1$, omitting those equal to $1$ and arrange the rest in ascending order (if they are all $1$, the inequality is trivial). $\endgroup$ – Quang Hoang Oct 28 '14 at 5:34

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