Maximum Sum Value

Let $S = a_1+a_2+a_3+...+a_{10}$ .

By AM-GM, we have $\displaystyle \sum_{k=1}^{10}\frac{k-a_k}{10}\ge \sqrt[10]{\prod_{k=1}^{10}(k-a_k)} = 1\Rightarrow$ $\displaystyle \sum_{k=1}^{10}(k-a_k)\ge 10 \Rightarrow$ $\displaystyle \frac{10\cdot 11}{2} - S\ge 10 \Rightarrow$ $55 - S\ge 10 \Rightarrow$ $S\le 45$

It is easy to show that this bound is achievable. Indeed, AM-GM tells us that equality only holds if all the addends are equal. This means that $1-a_1=2-a_2=3-a_3=...=10-a_{10}$ However, the product of these numbers is $1$ , so if all of them are equal, than each is equal to $1$ . Setting each value equal to $1$ , we find that $a_1=0$ $a_2=1$ $a_3=2$ $...$ $a_{10}=9$ and this works, since $S=0+1+2+...+9=45.$ Thus, the maximum possible value of $S$ is $\boxed{45}$

For brevity, we will denote the sum we want by $S.$

Consider the set $A = \{1-a_1, 2-a_2, ..., 10 - a_{10} \}.$ Its arithmetic mean is

$$\frac{1}{10} (1-a 1+2-a 2+...+10-a {10}) $$ $$= \frac{1}{10}[55-(a 1+a 2+...+a {10})] = \frac{55-S}{10}$$

and its geometric mean is $$\sqrt[10]{(1-a 1)(2-a 2)...(10-a_{10})}.$$

We are given that $(1-a_1)(2-a_2)...(10-a_{10}) \ge 1,$ thus $\sqrt[10]{(1-a_1)(2-a_2)...(10-a_{10})} \ge 1$ as well.

We now use the AM-GM inequality:

$\frac{55-S}{10} \ge \sqrt[10]{(1-a_1)(2-a_2)...(10-a_{10})},$ so $\frac{55-S}{10} \ge 1.$ Solving this inequality for $S$ gives $S \le 45.$ $S = 45$ is indeed achievable (let $a_1=0,a_2=1, ..., a_{10}=9$ ), so the maximum possible value is $\boxed{45}.$

Using AM-GM inequality for 10 positive reals, we have \­(1\le (1-a 1)(2-a 2) \cdots (10-a {10}) \le {\left( {\frac{{1 - {a 1} + 2 - {a 2} + \cdots + 10 - {a {10}}}}{{10}}} \right)^{10}}.\­)

Therefore, ${a_1} + {a_2} + \cdots + {a_{10}} \le 1 + 2 + \cdots + 9 = 45.$

For $a_1=0,\;\cdots,\;a_{10}=9$ , we will have the equality.

From the AM-GM inequality,

$\frac{(1-a_1)+(2-a_2).......(9-a_9)+(10-a_{10})}{10} \geq$ $((1-a_1)(2-a_2).......(9-a_9)(10-a_{10}))^{1/10} \geq 1$

$\displaystyle \Rightarrow \frac{55-\sum_{i=1}^{10}a_i}{10} \geq 1$

Rearranging and solving, $\sum_{i=0}^{10} a_i \leq 45$

Let {(1-\a 1), (2-\a 2),...,(10-\a 10)} be the set of numbers of interest. Using the AM-GM inequality, we see that (i) [(1-\a 1)\cdot (2-\a 2)\cdot...(10-\a 10)]^1/10 \leq [(1-\a 1)+ (2-\a 2)+...+(10-\a_10)]/10.

The right-hand side of the inequality can be written as [55-(\a 1+\a 2+...+\a 10)]/10. Multiplying -1 to both side of inequality reverses the inequality, thus yielding (ii) -[(1-\a 1)\cdot (2-\a 2)\cdot...(10-\a 10)]^1/10 \geq [(\a 1+\a 2+...+\a_10)-55]/10.

But it is said that (1-a 1)(1-a 2)...(1-a_10) is greater than or equal to 1, then its negative is less than or equal to 1. Then the maximum value of the left hand side of inequality (ii) is -(1)^10. Rewriting (ii) yields,

(iii) -1^10 \geq [(\a 1+\a 2+...+\a_10)-55]/10.

With some manipulation, we see that (\a 1+\a 2+...+\a 10) \leq 45. Therefore, the maximum value of \a 1+\a 2+...+\a 9+\a_10 is 45 .

As, the given equation is, $(1-a_1)(2-a_2)…(9-a_9)(10-a_10) \geq 1$

Applying AM-GM inequality for LHS we get, $(\frac{55-x}{10})^{10} \geq 1$ (Where 'x' is the required sum)

It gives, $\boxed{x=45}$

{ (1−a1)+(2−a2)+…+(9−a9)+(10−a10) } / 10 > = {(1−a1)(2−a2)…(9−a9)(10−a10)}^1/10 => 55 - ( a1+a2+...+a10 ) >= 1 => a1+a2+...+a10 <= 45

Because $a_i \leq i$ , we know that each of $1-a_1, 2-a_2,... 10-a_{10}$ are greater than or equal to 0. Since the product of $1-a_1, 2-a_2,... 10-a_{10}$ is not 0, we know that none of $1-a_1, 2-a_2,... 10-a_{10}$ can be equal to zero, thus they must all be simply greater than 0. Therefore, we can take the tenth root of both sides of the original inequality to get:

$1 \leq \sqrt[10]{(1-a_1)(2-a_2)(3-a_3)...(10-a_{10})}$

Note that the right side of the above equation is the geometric mean of 10 positive real numbers. By AM-GM, we know that the arithmetic mean of these 10 numbers is necessarily greater than their geometric mean. Therefore we have:

$1 \leq \sqrt[10]{(1-a_1)(2-a_2)(3-a_3)...(10-a_{10})}$

$\leq \frac{(1-a_1)+(2-a_2)+(3-a_3)+...+(10-a_{10})}{10}$

$= \frac{1+2+...+10-(a_1+a_2+...+a_{10})}{10}$

$= \frac{55-(a_1+a_2+...+a_{10})}{10}$

Thus we have the inequality:

$1 \leq \frac{55-(a_1+a_2+...+a_{10})}{10}$

Rearranging terms, we have the answer we are looking for:

$a_1+a_2+...+a_{10} \leq 45$

Applying AM-GM to $x_i = i-a_i\geq 0$ , we have $\frac{(1-a_1)+(2-a_2)+\ldots+(9-a_9)+(10-a_{10})}{10} \geq \sqrt[10]{(1-a_1)(2-a_2)\ldots(9-a_9)(10-a_{10})} \geq 1$ . Thus, $55 - (a_1 + a_2 + \ldots + a_9 + a_{10}) \geq 10$ $\Rightarrow a_1 + a_2 + \ldots + a_9 + a_{10} \leq 45$ . This can be achieved by setting $a_i = i-1$ .

$\large \ let \ A = a_1 +a_2 +...a_10$ $\large \ Using \ the \ AM-GM \ inequality:$ $\frac{(1-a_1) + (2-a_2) +...+(10-a_{10})}{10}\geq\ [(1-a_1)(2-a_2)\cdots\ (10-a_{10})]^{\frac{1}{10}} = 1$ Now the maximum value of A is achieved when equality holds so multiplying by 10 and rearranging the terms: $(1+2+..+10) - A = 10 \Rightarrow\ 55 - A = 10 \therefore\ A = 45$ $\large \ equality \ holds \ when \ a_i = i-1$

( $\ note: \ equality \ can \ hold \ in \ other \ cases$ )

Here, the series is greater than or equal to 1

The maximum value of ai can be i

When ai=i then the given series becomes 0

So, the maximum possible value of ai will be 1 less than i

So the required maximum values of ai will be

a1=0 a2=1 a3=2 a4=3 a5=4 a6=5 a7=6 a8=7 a9=8 a10=9

The maximum value

a1+a2+a3+a4+a5+a6+a7+a8+a9+a10 = 0+1+2+3+4+5+6+7+8+9 = 45

                                      Answer: 45

(a 1,a 2,a 3,...,a 10) = (0,1,2,3, ...,9), max(sum) = (9 * 10) / 2 = 45