Product of $\left(1-\frac{x}{k}\right)$ with or without uniform $k$

Question

Is it possible to prove this: $$\left(1−\frac{x}{k}\right)^k\ge(1−A_1)\cdot(1−A_2)\cdots(1−A_k)$$ for x>0, and k≥1, where k is a whole number. where $$A_1+A_2+...+A_k = x$$ At least one number in $A_1, A_2,...,A_k$ does not equal to the other.

Answer 1

$$\left(1-\frac{x}{k}\right)^k~\geq~(1-A_1)(1-A_2)\cdots(1-A_k)$$

By using the condition $A_1+A_2+ \cdots + A_k~=~x$ the inequality becomes

$$\begin{align} (1-A_1)(1-A_2)\cdots(1-A_k)~&\leq~\left(1-\frac{(A_1+A_2+ \cdots + A_k)}{k}\right)^k\\ &\leq~\left(\frac{k-(A_1+A_2+ \cdots + A_k)}{k}\right)^k\\ &\leq~\left(\frac{(1-A_1)+(1-A_2)+ \cdots + (1-A_k)}{k}\right)^k\\ \sqrt[\huge{k}]{(1-A_1)(1-A_2)\cdots(1-A_k)}~&\leq~\frac{(1-A_1)+(1-A_2)+ \cdots + (1-A_k)}{k} \end{align}$$

The L.H.S. of this inequality is just the geometric mean of the $k$-factors $(1-A_k)$ and the R.H.S. is the arithmetic mean of the $k$-summands $(1-A_k)$. The GM-AM inequalitiy is known to be true and so the given is.