+ 1 all terms then amgm

MortemEtInteritum wrote:

Should be $2^{k-1}$

Let $S_i=1+a_1+\cdots a_i$. Now, we have the summation, $\sum_{i=0}^{n-1} \frac{S_{i+1}-S_i}{S_i}=\sum_{i=0}^{n-1} \frac{S_{i+1}}{S_i}-n\le n(\sqrt[n]{\frac{S_{n}}{n}})-n=n$ and this is clearly achievable when $a_i=2^{i-1}$.

Add an extra term $a_0 = 1$. Let $S_n$ denote the sum of the terms of $a_i$ until $a_n$. So, $S_{n} = 2^n$ $\frac{a_1}{1} + \frac{a_2}{1+a_1} + ... + \frac{a_n}{1 + a_1 + a_2 + ... + a_{n-1}}$ $= \frac{a_1}{1} + 1 + \frac{a_2}{1+a_1} + 1 + ... + \frac{a_n}{1 + a_1 + a_2 + ... + a_{n-1}} + 1 - n$ $= \frac{S_1}{S_0} + \frac{S_2}{S_1} + ... + \frac{S_n}{S_{n-1}} - n$ $\ge n \sqrt[n]{S_n} - n$ by AM-GM $= 2n - n$ $= n$ So, the minimum value is $n$, which is achieved when $a_k = 2^{k-1}$

pretty easy for national MO tho

qwqwq wrote: pretty easy for national MO tho feels more like regional

MortemEtInteritum wrote: Let $n\geq 2$ be some fixed positive integer and suppose that $a_1, a_2,\dots,a_n$ are positive real numbers satisfying $a_1+a_2+\cdots+a_n=2^n-1$. Find the minimum possible value of $$\frac{a_1}{1}+\frac{a_2}{1+a_1}+\frac{a_3}{1+a_1+a_2}+\cdots+\frac{a_n}{1+a_1+a_2+\cdots+a_{n-1}}$$ After Applying Chebeshev 2 times I got minimum value $$ \frac{2}{2^{n} 3 -1}$$ EDIT : It's wrong because I assumed ${a_i}$ to be ordered similarly..

MatBoy-123 wrote: After Applying Chebeshev 2 times I got minimum value $$ \frac{2}{2^{n} 3 -1}$$ Is it achievable? qwqwq wrote: pretty easy for national MO tho Early Canada MO problems are, in my experience, like that. The hardest part about the contest as a whole is the short time limit.

yikes, it took me quite a while to solve this problem- once you see the solution, its a one-liner. @IAmTheHazard yeah- I definitely agree. you only get 3 hours to write the contest. :O

Cause n≥2 can we say that the minimum value is 2?

SpoonLoveMath wrote: Cause n≥2 can we say that the minimum value is 2? No, it says possible. 2 is not possible for n>2. Otherwise you could just say negative infinity.

Being the only one who used Induction on $n$ and bashed for 40 minutes be like:

After adding one to all terms, AM-GM yields $$\sum_{i=1}^n\left(\frac{a_i}{\sum_{j=0}^{i-1}a_j}+1\right)-n=\sum_{i=1}^n\left(\frac{\sum_{j=0}^i a_j}{\sum_{j=0}^{i-1}a_j}\right)-n\ge n\left(\frac{\sum_{j=0}^i a_j}{\sum_{j=0}^{i-1}a_j}\right)^{\frac{1}{n}}-n=2n-n=n,$$where we let $a_0=1.$ Equality holds when $a_i=2^{i-1}$ for $1\le i\le n.$ $\square$

awesomeming327. wrote: Being the only one who used Induction on $n$ and bashed for 40 minutes be like: Induction is a possible method. Actually I did it with induction: Quote: Prove that \[\frac{a_1}{1}+\frac{a_2}{1+a_1}+\frac{a_3}{1+a_1+a_2}+\cdots+\frac{a_n}{1+a_1+a_2+\cdots+a_{n-1}}\ge n\sqrt[n]{1+a_1+\cdots+a_n}-n.\] I get this totally by finding the pattern... Afterwards, I found that the two methods both use AM-GM, and in a rather similar way.

The answer is $\boxed{n}$, achieved by $a_i=2^{i-1}$. For the proof, note that \begin{align*} \;& \frac{a_1}{1}+\frac{a_2}{1+a_1}+\frac{a_3}{1+a_1+a_2}+\dots+\frac{a_n}{1+a_1+a_2+\cdots+a_{n-1}} \\ =\;& \frac{1+a_1}{1}+\frac{1+a_1+a_2}{1+a_1}+\frac{1+a_1+a_2+a_3}{1+a_1+a_2}+\dots+\frac{1+a_1+a_2+\dots+a_n}{1+a_1+a_2+\dots+a_{n-1}}-n \\ \ge\;& n\sqrt[n]{1+a_1+a_2+\dots+a_n}-n \\ =\;& n\;\blacksquare \end{align*}

Since, \[ a_1+a_2+\cdots+a_n=2^n-1 \] Adding +1 for every fraction in given expression, $\frac{a_1}{1}+ 1 + \frac{a_2}{1+a_1} + 1 + \frac{a_3}{1+a_1+a_2} + 1 +\cdots+\frac{a_n}{1+a_1+a_2+\cdots+a_{n-1} } + 1 = \frac{a_1 + 1}{1} + \frac{a_2 + a_1 + 1}{1 + a_1} + \frac{a_3 + a_2 + a_1 + 1}{1 + a_1 + a_2} + \cdots + \frac{a_1+a_2+\cdots+a_n + 1}{1+a_1+a_2+\cdots+a_{n-1}} $ By AM-GM, \begin{align*} \frac{a_1}{1}+\frac{a_2}{1+a_1}+\frac{a_3}{1+a_1+a_2}+\cdots+\frac{a_n}{1+a_1+a_2+\cdots+a_{n-1}} & \geq n \left( \sqrt[n]{\frac{a_1 + 1}{1} \cdot \frac{a_2 + a_1 + 1}{1 + a_1} \cdot \frac{a_3 + a_2 + a_1 + 1}{1 + a_1 + a_2} \cdot \cdots \cdot \frac{a_1+a_2+\cdots+a_n + 1}{1+a_1+a_2+\cdots+a_{n-1}}} \right) - n \\ & \geq 2n - n \\ & \geq n \\ \end{align*} $\blacksquare $

Define $b_i = 1 + a_1 + a_2 + \cdots a_i$ and add $1$ to each of the terms in the quantity we want to minimize. We have the quantity, $b_1 +\frac{b_2}{b_1} + \cdots \frac{b_n}{b_{n-1}} - n$ and now AM-GM clearly implies that this quantity is greater than equal to $n$. For the construction, consider $a_i = 2^{i-1}$.

One-liner: \begin{align*}\sum_{i = 1}^n \frac{a_i}{1 + a_1 + a_2 + \dots + a_{i - 1}} = \sum_{i = 1}^n \frac{1 + a_1 + a_2 + \dots + a_i}{1 + a_1 + a_2 + \dots + a_{i - 1}} - n \leq n \sqrt[n]{1 + a_1 + a_2 + \dots + a_n} - n = n.\end{align*}