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This question already has an answer here:

I am doing past exam papers preparing for the finals and I came across this questions about three times:

Prove that:

$$\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\geq \sqrt[n]{a_{1}.a_{2}...a_{n}}$$ Different papers require different approaches to proving it but most of them use induction. Can someone please explain to me how to prove it using induction? I can do the first steps but get stuck at proving that:

$$\frac{a_{1}+a_{2}+\cdots+a_{k}+a_{k+1}}{k+1}\geq \left ( \frac{a_{1}+a_{2}+\cdots+a_{k}}{k} \right )^{\frac{1}{k+1}}\left ( a_{k+1} \right )^{\frac{1}{k+1}}$$

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marked as duplicate by user147263, TravisJ, user99914, David K, Najib Idrissi May 23 '15 at 14:47

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    $\begingroup$ "Can someone please explain to me how to prove it using induction?" The WP page does it rather nicely, providing four different proofs (the one using forward-backward-induction is a gem). $\endgroup$ – Did Aug 15 '13 at 10:01
  • $\begingroup$ Thank you very much! I will go through these proves and understand them. $\endgroup$ – please delete me Aug 15 '13 at 10:28
  • $\begingroup$ You can find some proofs in the question linked from A question on mean value inequality. $\endgroup$ – Martin Sleziak Aug 15 '13 at 13:09
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Here is a simple proof. It isn't an induction, but can be converted to such a method, which may serve.

Let the geometric mean (right hand side of your expression) equal $G$ and the arithmetic mean equal $A$. If the $a_i$ are not all equal, then there is a largest $a_1$ and a smallest $a_n$ (say - arranging in order of magnitude) so that $$a_1\gt G \gt a_n$$ and therefore $(G-a_1)(G-a_n)\lt 0$

Now consider the effect of replacing $a_1$ with $b_1=G$ and $a_n$ with $b_n=\frac {a_1a_n}G$ and $b_i=a_i$ otherwise so that $b_1 b_n=a_1a_n$ and the geometric mean is unchanged, but we have $$a_1+a_n\gt b_1+b_n=\frac {G^2+a_1a_n}{G}$$ because (multiplying by $G\gt 0$ and gathering terms) $$0 \gt G^2+a_1a_n-G a_1-G a_n=(G-a_1)(G-a_n)$$

If we let $B$ be the arithmetic mean of the $b_i$ we have $A\gt B$. After at most $n-1$ steps of the same kind, all the original $a_i$ will have been replaced by $G$ and we get something like:$$A\gt B\gt B_2\gt\dots\gt B_{n-1}=G$$

If you need an induction, you can do an induction based on this method. Induct on the number of the $a_i \neq G$ noting that the number can't be $1$ so a little care is needed getting started. If we have proved for $r$ of the $a_i\neq G$ then we take a case with $r+1$ and reduce to the case of $r$ using the method above. This also easily gives the case of equality.

This does not give an induction on $n$, of course - so whether it would be a valid solution would depend on the wording of the question.

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Let me discuss your doubt only. Complete proof you have in your notes. As you have written in your question I shall discuss the particular step you did nor understand.

$$\frac{a_1 + a_2 + \dots + a_k + a_{k+1}}{k+1} = \frac{k \frac{a_1 + a_2 + \dots + a_k}{k} + a_{k+1}}{k+1} \geq (\frac{a_1 + a_2 + \dots + a_k}{k})^{\frac{k}{k+1}}(a_{k+1})^{\frac{1}{k+1}}$$

Here we have taken $k$-times $\frac{a_1 + a_2 + \dots + a_k}{k}$ and $1$-time $a_{k+1}$.

Now applying the previous step of your induction you shall get

$$\frac{a_1 + a_2 + \dots + a_k}{k} \geq (a_1 a_2 \dots a_k)^{\frac{1}{k}}$$

Putting this value in the first relation you shall get

$$\frac{a_1 + a_2 + \dots + a_{k+1}}{k+1} \geq (a_1 a_2 \dots a_{k+1})^{\frac{1}{k+1}}$$

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  • $\begingroup$ The first two fractions are not equal. Did you want to write $k \frac{a_1 + a_2 + \dots + a_k}{k} + a_{k+1}$ instead of $k \frac{a_1 + a_2 + \dots + a_k}{k}$? Also mistake you your question look like a typo. $\endgroup$ – Martin Sleziak Aug 15 '13 at 13:03
  • $\begingroup$ @Sleziak Thank you. I did not check it. It is as you are saying. $\endgroup$ – Dutta Aug 15 '13 at 13:35

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