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

*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 :

*

*Have I solved the questions correctly?

*How to prove $\sum{a_1}\cdot{a_2}\cdot{a_3}\ge \frac{n(n-1)(n-2)}{3!}G^3$
Thanks

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.
