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I want to know how to prove this inequality by mathematical induction: $a_k's$ are nonnegative numbers. Prove that$$a_1a_2\cdots a_n\leq \left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^n.$$ In the inductive step, I tried using the inequality $ab\leq \frac{k}{k+1}a^\frac{k+1}{k}+\frac{1}{k+1}b^{k+1}$ but I got over estimate of what I was looking for.

Is there a better way of proving this by induction?

Michael Hardy
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OKPALA MMADUABUCHI
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    there is a proof by induction here https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Proof_by_induction. I'm not sure if you need to use induction, but there are (in my opinion) simpler proofs (page 48 here http://people.whitman.edu/~gordon/higher_math.pdf) – pancini Aug 24 '15 at 17:28
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    Can we assume $a^tb^{1-t} \le ta + (1-t)b$ for $a,b\ge 0, t \in [0,1]?$ – zhw. Aug 24 '15 at 17:29
  • @Elliot I know other methods of proving this of course. However, I need to see to prove it by induction. – OKPALA MMADUABUCHI Aug 24 '15 at 17:35
  • @zhw I have said it already that that assumption didnt give me the desired result! – OKPALA MMADUABUCHI Aug 24 '15 at 17:35
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    Assuming the inequality works for $n,$ we have

    $$(a_1a_2\cdots a_{n+1})^{1/(n+1)}=[(a_1a_2\cdots a_n)^{1/n}]^{n/(n+1)}a_{n+1}^{1/(n+1)}$$ $$ \le \frac{n}{n+1}(a_1a_2\cdots a_n)^{1/n} + \frac{1}{n+1}a_{n+1}.$$Now use the induction hypothesis.

     – zhw. Aug 24 '15 at 17:51
  • @zhw : You're also assuming it works for $2$. ${}\qquad{}$ – Michael Hardy Aug 24 '15 at 18:48
  • @MichaelHardy Right, but he said we can assume that; see comments above. – zhw. Aug 24 '15 at 18:49

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Maybe this is one of those cases in which an induction proof is easier if one makes the statement to be proved stronger, because then one has a stronger induction hypothesis to use. The inequality in the question is $$ a_1^{1/n}\cdots a_n^{1/n} \le \frac 1 n \left(a_1 + \cdots+ a_n\right). $$ These are a geometric mean and an arithmetic mean of $n$ numbers with equal weights $1/n$. More generally one has weights $p_1,\ldots,p_n\ge 0$, $p_1+\cdots+p_n=1$. The inequality is then $$ a_1^{p_1} \cdots a_n^{p_n} \le p_1 a_1 + \cdots + p_n a_n. $$

Now let's do the induction step: \begin{align} a_1^{1/(n+1)} \cdots a_{n+1}^{1/(n+1)} & = \Big(a_1^{1/n} \cdots a_n^{1/n}\Big)^{n/(n+1)} \Big( a_{n+1} \Big)^{1/(n+1)} & \text{(weights $\frac n {n+1}$ and $\frac 1 {n+1}$)} \\[10pt] & \le \frac n {n+1} \Big(a_1^{1/n} \cdots a_n^{1/n}\Big) + \frac 1 {n+1} a_{n+1} \\ & {}\qquad \text{(by the induction hypothesis in case 2)} \\[12pt] & \le \frac n {n+1} \Big( \frac 1 n \left( a_1+\cdots+a_n \right) \Big) + \frac 1 {n+1} a_{n+1} \\ & {} \qquad \text{(by the induction hypothesis in case $n$)} \\[10pt] & = \frac 1 {n+1} \left(a_1 + \cdots + a_{n+1} \right). \end{align}

Michael Hardy
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