# A Beautiful Problem

For some $$m\in\mathbb N$$, there are $$2^m$$ positive real numbers written on a board product of which is $$1$$. At each step, we can choose two numbers $$a,b$$ and replace both by $$a+b$$. Prove that after $$2^{m-1}m$$ steps the sum of all numbers on board will be at least $$4^m$$.

Note by Jubayer Nirjhor
3 years, 11 months ago

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How does this work? I'm confused by the #WishfulThinking tag. Is the question correct?

The sum of all the numbers is invariant and equal to $$2^m$$ and can never be $$4^m$$

. Also, the number of numbers on the boards decreases by one every step. So after $$2^m -1$$ steps, there should only be $$1$$ number left on the board, after which no moves are possible. So you can't have $$2^{m-1}*m$$ moves since $$2^{m-1}*m > 2^{m} -1$$ for $$m \geq 2$$

- 3 years, 11 months ago

We replace "both" of $$a,b$$ by $$a+b$$ so $$(a,b)$$ changes to $$(a+b,a+b)$$. Thus the number of terms remains same.

- 3 years, 11 months ago

Ah. Okay. The question is still ambiguously worded though. A better wording would be " we replace each of $$a,b$$ with $$(a+b)$$"

Anyways, the problem seems easy. Just apply the "step" to the two largest numbers which has a sum of $$\geq 2$$. Each "step" doubles the sum of the numbers. Initial sum = $$a+b$$, final sum $$= 2(a+b)$$. Applying it $$2^{m-1}m$$ times, the sum of those two numbers is now $$\geq 2^{2^{m-1}m}$$. which is obviously greater than $$2^{2m}$$

- 3 years, 11 months ago

Not so fast! If initial terms are $$a_1,...,a_{2^m}$$ then initial sum is $$a_1+\cdots+a_{2^m}$$, while the sum after applying the step is $$2a_1+2a_2+a_3+\cdots+a_{2^m}$$. So the sum is not doubled.

- 3 years, 11 months ago

Sorry. I wasn't clear. I meant the sum of the largest two numbers when I said the sum will be doubled.

My solution was to just ignore the other numbers. Take only the two largest numbers. Their sum has to $$\geq 2$$. Now, one step doubles the sum of these two numbers. Applying the step $$2^{m-1}*m$$ times we see that the sum of these two numbers is now $$\geq 2^{2^{m-1}m}$$. The sum of the numbers we ignored is $$\geq 0$$. so doesn't matter. Clearly, the sum of all the numbers is now $$2^{2^{m-1}m}$$ which is clearly greater than $$2^{2m}$$.

- 3 years, 11 months ago

You don't decide which numbers to choose. The statement holds for any order.

- 3 years, 11 months ago

Any order? No it doesn't. Take the case where $$m = 2$$. Take $$x_1, x_2, 2-x_1,2-x_2$$. Sum of these $$2^2$$ numbers is $$2^2$$

Apply the step 4 times on $$x_1, x_2$$. Then the four numbers are $$4(x_1+x_2),4(x_1+x_2), 2-x_1, 2-x_2$$.

Sum of these numbers is $$4 + 7(x_1+x_2)$$. If we take $$x_1,x_2$$ arbitrarily small, it will not be greater than $$16$$.

- 3 years, 11 months ago

Sorry. A change has been made. The initial product of the numbers is $$1$$. No restriction on the sum.

- 3 years, 11 months ago