## Need help

For students of class 11-12 (age 16+)
sharod23
Posts: 4
Joined: Fri Aug 22, 2014 1:45 am

### Need help

How to prove that the product of n numbers is maximum when and only when all the numbers are equal provided that their sum is a constant? (n is an arbitrary natural number)

Rabeeb
Posts: 25
Joined: Tue Dec 14, 2010 7:52 pm
Contact:

### Re: Need help

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

sharod23
Posts: 4
Joined: Fri Aug 22, 2014 1:45 am

### Re: Need help

Wow, I am impressed. Thank you.

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 $a_1=a_2=\cdots=a_n$
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.