## Need help

### Re: Need help

I think the ques. shud be "the product of n

Let us suppose the statement to be P(n) for natural number n.

We will show the following:

P(2) is true, P(n) is true implies P(2n) is true, P(n+1) is true implies P(n) is true.

P(2) is trivial.

Let P(n) is true. For P(2n), we take a: Sum = 2na. For every element = a, product = (a^2)^n

For unequal elements, we divide them into two groups each of n elements. At least two unequal elements are present and we take them in the first n elements.

If sum of 1st n elements = sum of 2nd n elements = na, we have,

product of all 2n elements = product of 1st n elements * product of 2nd n elements < (a^n) * (a^n) = (a^2)^n

Again if sum of 1st n elements = na - d', we take d: d' = nd (d', d not equal to zero)

Sum of 1st n elements = n(a-d), sum of 2nd n elements = n(a+d)

Now product of 1st n elements * product of 2nd n elements < (a-d)^n * (a+d)^n = (a^2 - d^2)^n < (a^2)^n

So max product is obtained only when all elements are equal, i.e, P(n) implies P(2n)

Now let P(n+1) is true. For P(n), we take a: Sum = na.

Config.1(of n elements): all elements are equal. Add a new element a to it get Config. 1* of n+1 elements

Config. 2(of n elements): 2 or more elements are unequal. Add a new element a to it to get Config. 2* of n+1 elements

Note that both config. 1* and config. 2* have last element a.

let, product(config. k) = product of elements of config. k

By hypothesis[P(n+1) is true], product(config. 1*) > product(config. 2*)

Dividing by a, product(config. 1) > product(config. 2), so P(n) is true if P(n+1) is true.

So by induction, the result immediately follows...

**non-negative**numbers is maximum when and only when all the numbers are equal provided that their sum is a constant"Let us suppose the statement to be P(n) for natural number n.

We will show the following:

P(2) is true, P(n) is true implies P(2n) is true, P(n+1) is true implies P(n) is true.

P(2) is trivial.

Let P(n) is true. For P(2n), we take a: Sum = 2na. For every element = a, product = (a^2)^n

For unequal elements, we divide them into two groups each of n elements. At least two unequal elements are present and we take them in the first n elements.

If sum of 1st n elements = sum of 2nd n elements = na, we have,

product of all 2n elements = product of 1st n elements * product of 2nd n elements < (a^n) * (a^n) = (a^2)^n

Again if sum of 1st n elements = na - d', we take d: d' = nd (d', d not equal to zero)

Sum of 1st n elements = n(a-d), sum of 2nd n elements = n(a+d)

Now product of 1st n elements * product of 2nd n elements < (a-d)^n * (a+d)^n = (a^2 - d^2)^n < (a^2)^n

So max product is obtained only when all elements are equal, i.e, P(n) implies P(2n)

Now let P(n+1) is true. For P(n), we take a: Sum = na.

Config.1(of n elements): all elements are equal. Add a new element a to it get Config. 1* of n+1 elements

Config. 2(of n elements): 2 or more elements are unequal. Add a new element a to it to get Config. 2* of n+1 elements

Note that both config. 1* and config. 2* have last element a.

let, product(config. k) = product of elements of config. k

By hypothesis[P(n+1) is true], product(config. 1*) > product(config. 2*)

Dividing by a, product(config. 1) > product(config. 2), so P(n) is true if P(n+1) is true.

So by induction, the result immediately follows...

### Re: Need help

Wow, I am impressed. Thank you.

- Phlembac Adib Hasan
**Posts:**1016**Joined:**Tue Nov 22, 2011 7:49 pm**Location:**127.0.0.1-
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### Re: Need help

So, for your problem, let the numbers are $a_1,...,a_n$ and suppose $M=a_1+a_2+\cdots+a_n$. We'll get $\dfrac M n\ge \sqrt[n]{a_1a_2\cdots a_n}\Longrightarrow\left (\dfrac M n\right )^n\ge a_1a_2\cdots a_n$. The maximum is achieved when all the numbers are equal.AM-GM inequality:

For positive real numbers $a_1, a_2,\cdots ,a_n$, we have\[AM=\frac{a_1+a_2+\cdots+a_n}n\ge \sqrt[n]{a_1a_2\cdots a_n}=GM\]with equality when $a_1=a_2=\cdots=a_n$

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