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$A$ and $G$ are arithmetic mean and the geometric mean respectively of $n$ positive real numbers $a_1$,$a_2$,$\ldots$,$a_n$ . Prove that

  1. $(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
  2. if $k$$\gt$$0$, $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$$\ge$$(k+G)^n$

My try (Ques -1): To prove $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$

$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)=1+\sum{a_1}+\sum{a_1}\cdot{a_2}+\sum{a_1}\cdot{a_2}\cdot{a_3}+\cdots+(a_1\cdot{a_2}\cdots{a_n})$

Consider $a_1,a_2,\ldots,a_n$ be $n$ positive real numbers and then apply AM$\ge$GM

$\frac{(a_1+a_2+\cdots+a_n)}{n}\ge(a_1\cdot{a_2}\cdots{a_n})^\frac{1}{n}$ imples $\sum{a_1}\ge n\cdot{G}$ because $G=(a_1\cdot{a_2}\cdots{a_n})^\frac{1}{n}$

again consider $(a_1\cdot{a_2}),(a_1\cdot{a_3}),\ldots({a_1}\cdot{a_n}),(a_2\cdot{a_3}),(a_2\cdot{a_4})\ldots,(a_2\cdot{a_n}),(a_3\cdot{a_4})\ldots(a_{n-1}\cdot{a_n})$ be $\frac{n(n-1)}{2!}$ positive real numbers.

Then Applying AM$\ge$GM , we get, $\frac{\sum{a_1}\cdot{a_2}}{\frac{n(n-1)}{2!}}$ $\ge$ $\bigl(a_{1}^{n-1}\cdot{a_{2}^{n-1}}\cdots{a_{n}^{n-1}}\bigl)^\frac{2!}{n(n-1)}$ implies $\sum{a_1}\cdot{a_2}\ge \frac{n(n-1)}{2!}G^2$

Similary , $\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$ and so on.

Therefore, $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)\ge 1+nG+\frac{n(n-1)}{2!}G^2+\frac{n(n-1)(n-2)}{3!}G^3+\cdots+G^n$

Therefore, $(1+a_1)\cdot(1+a_2)\cdots(1+a_n)\ge (1+G)^n$

To prove $(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$

Consider $(1+a_1),(1+a_2),\ldots,(1+a_n)$ be $n$ positive real numbers and then applying AM$\ge$GM

$\frac{(1+a_1)+(1+a_2)+\cdots+(1+a_n)}{n} \ge [(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)]^\frac{1}{n}$

$\frac{n+a_1+a_2+\cdots{a_n}}{n}\ge [(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)]^\frac{1}{n}$

$(1+A)^n\ge(1+a_1)\cdot(1+a_2)\cdot\cdots(1+a_n)$

My try (Ques -2): To prove $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$

Consider $(k+a_1),(k+a_2)\ldots(k+a_n)$ be $n$ positive real numbers and applying AM$\ge$GM

$\frac{(k+a_1)+(k+a_2)+\cdots(k+a_n)}{n}\ge[(k+a_1)\cdot(k+a_2)\cdots(k+a_n)]^\frac{1}{n}$

$(k+A)^n\ge (k+a_1)\cdot(k+a_2)\cdots(k+a_n)$

To prove $(k+a_1)\cdot(k+a_2)\cdots(k+a_n)$$\ge$$(k+G)^n$

$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)=k^n+k^{n-1}\sum{a_1}+k^{n-2}\sum{a_1\cdot{a_2}}+\cdots+(a_1\cdot{a_2}\cdots{a_n})$

now

$\sum{a_1}\ge n\cdot{G}$,

$\sum{a_1}\cdot{a_2}\ge \frac{n(n-1)}{2!}G^2$

$\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$ and so on.

Therefore

$(k+a_1)\cdot(k+a_2)\cdots(k+a_n)\ge k^n+nk^{n-1}G+\frac{n(n-1)}{2!}k^{n-2}G^2+\cdots+G^n$

Therefore, $(k+a_1)\cdot(k+a_2)\cdots(k+a_n)\ge (k+G)^n$

My Ques :

  1. Have I solved the questions correctly?
  2. How to prove $\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$

Thanks

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  • $\begingroup$ For 2: what is the geometric mean if the terms on the left. $\endgroup$
    – richrow
    Commented Jul 31, 2021 at 8:56
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    $\begingroup$ Fyi Huygen's Inequality. $\endgroup$
    – rtybase
    Commented Jul 31, 2021 at 9:24
  • $\begingroup$ Note: Another approach is to use Forward-Backward induction, since the $ n =2$ case is obvious/easy. $\endgroup$
    – Calvin Lin
    Commented Jul 31, 2021 at 18:31

3 Answers 3

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I think there is a much faster way.

  1. $\log(k+x)$ is concave, hence by Jensen's inequality $\log(k+A)\geq\frac{1}{n}\sum_{m=1}^{n}\log(k+a_m)$.
    Exponentiation leads to $(k+A)^n \geq \prod_{m=1}^{n}(k+a_m)$.
  2. Let $b_m=\log(a_m)$ or, equivalently, $a_m=e^{b_m}$. We have that $\log(k+e^x)$ is convex, since $\frac{d^2}{dx^2}\log(k+e^x)=\frac{ke^x}{(k+e^x)^2}$. This leads to $(k+G)^n\leq \prod_{m=1}^{n}(k+a_m)$.
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Your approach works fine. I feel I should point out that the second result follows trivially from the first one by thinking about what happens when you multiply all the $a_i$ by $k$. Also, there are a few places you say "integer" when you mean real number.

For the missing step in your method, you can use the same approach you did for the sum of products of $2$ variables for the sum of products of $m$ variables.

To do this, you simply need to count how many terms there are in the sum, which as you have noticed is $n \choose m$, and then (in order to compute the GM of these terms) you need to know how many of these terms the variable $a_i$ appears in. How could you count these? Hint: What are the other variables appearing in a term containing $a_i$?

Then you can simplify the exponent to get the answer you want.

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  • $\begingroup$ As @rtybase points out, there are other shorter approaches too! But they're a bit harder to spot. $\endgroup$ Commented Jul 31, 2021 at 9:43
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Adding on to @notallwrong's answer: For the case $n = 3$, you want to know how many times say $a_1$ appears. We have to make triples. Choose $a_1$. The remaining two can be chosen in $\binom{n-1}{2}$ ways.

For the case $n = 4$, you want to know how many times say $a_1$ appears. We have to make quadruples. Choose $a_1$. The remaining three can be chosen in $\binom{n-1}{3}$ ways. and so on.

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