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Let $n \ge 2$ be a positive integer and let $a_1, a_2, ... a_n$ be positive numbers such that $$ a_1\le a_2, a_1+a_2\le a_3, a_1+a_2+a_3\le a_4, ... ,a_1+a_2+...+a_{n-1}\le a_n$$ prove that $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{n-1}}{a_n} \le \dfrac{n}{2} \hspace{2cm} (1)$$ When does the equality holds?

Solution: I proceed as follows using Mathematical Induction.

For $n=2, \frac{a_1}{a_2} \le 1$. Let the (1) be true for $n=k$ i.e $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k} \le \dfrac{k}{2} \hspace{2cm} (2)$$ We need to prove $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k+1}{2} $$ Consider $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k}+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k}{2} +\dfrac{a_{k}}{a_{k+1}} \hspace{2cm} (3)$$ Since $$a_1+a_2+...+a_{k-1}+a_{k}\le a_{k+1} \hspace{2cm} (4)$$ also $$a_1+a_2+...+a_{k-1}\le a_{k} \hspace{2cm} (5)$$ Using 5 in 4, we get $$ a_{k}+a_{k}\le a_{k+1}$$ $$\dfrac{a_{k}}{a_{k+1}} \le \dfrac{1}{2}$$ using in (3), we get $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k}+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k+1}{2}$$ Is the procedure is correct. And when the equality holds... Thanks for any assistance

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    Your proof looks good. As to the question about when equality holds, the clues are right there in your proof. How big can $\frac{a_1}{a_2}$ be? Can you force equality? What about $\frac{a_2}{a_3}, \frac{a_3}{a_4},\ldots$ ? – 2017-01-07
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    How do you get $a_k+a_k \leq a_{k+1}$ from (4) and (5)? – 2017-01-07
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    Oops -- I fell in the same trap that yasir fell in -- ignore my first comment. Instead, look at rtybase's objection. – 2017-01-07
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    @rtybase Subtracting (5) from (4) – 2017-01-07
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    @MattG88 I want to see that :) – 2017-01-07
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    Sorry about my flawed comment -- low on sleep – 2017-01-07
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    OK, then how can I bring $\frac{a_{k+1}}{a_k} \le \frac{1}{2}$, and when the equality holds. – 2017-01-07
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    @quasi subtracting $$a_1+a_2+...+a_{k-1}\leq a_{k}$$ from $$a_1+a_2+...+a_k\leq a_{k+1}$$ implies $$-a_1-a_2-...-a_{k-1}\geq -a_{k}$$ which leads to $$a_1+a_2+...+a_{k-1}\leq \frac{a_{k+1}}{2}$$ – 2017-01-07
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    @rtybase ahah ok ;-) – 2017-01-07
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    wha? the step seems fine to me. It's a given that $a_1 +..... + a_n \le a_{n+1}$ for all $n$ then $a_1 + ..... + a_n + a_{n+1} \le a_{n+1} + a_{n+1} = 2a_{n+1} $ for all $n$. Admittedly it's a weird condition to be given but it was given. At that point one doesn't actually need induction. On can just point out $\sum_{i=1}^n \frac {a_{i}}{a_{i+1}} \le \sum_{1=i}^n \frac 12 = \frac n2$. – 2017-01-07
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    Equality holds when all the given inequalities are equalities. For example, the sequence 1,1,2,4,8,16,... – 2017-01-07
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    @rtybase -- you can't subtract inequalites that are in the same direction. You can add them, but you can't subtract them. A common illusion (which I fell victim to as well, even though I know better). – 2017-01-07
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    I'm not sure if an inductive proof will work, but if so, the induction can't use yasir's strategy. He tried to prove that $\frac{a_k}{a_{k+1}} \le \frac{1}{2}$, but that attempt is doomed -- it can be greater than $\frac{1}{2}$. For example, consider the 3 term sequence 1,2,3. – 2017-01-07
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    I wouid start by proving it for $k = 3$ (i.e., prove $$\frac{a_1}{a_2} + \frac{a_2}{a_3} \le \frac{3}{2}$$ For $k = 3$, it's provable, but my proof for the case $k = 3$ is not an "inductive type" proof. – 2017-01-07
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    Looking back at my comment where I prefixed it with @rtybase, that should instead have been mattG88. – 2017-01-07
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    Any other suggested solution? Thanks – 2017-01-09
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    @yasir: I have an outline (written down on paper) of a possible solution, but a fair amount of work would be needed to expand it to an actual proof. I'll do it when I get a chance (sometime this week), and I'll post it if it "survives" the expansion, unless someone else posts a similar solution (or a better one) first. – 2017-01-10
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    @yasir: Can I ask where you got the problem? – 2017-01-10
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    I wonder if the [HM-GM inequality](https://en.wikipedia.org/wiki/Pythagorean_means) yields anything useful here – 2017-01-11
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    I incorrectly use the result, Using 5 in 4, we get $a_{k}+a_{k}\le a_{k+1}$, so I have problem here, how to complete the proof to bring $\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k}+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k+1}{2}$ – 2017-01-15

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The idea is "local adjustments". If $k$ is the smallest integer such that $a_1+\cdots+a_{k-1} < a_k$, we adjust up $a_1$, $\cdots$, $a_{k-1}$ with a scale factor of $s=\frac{\sum_{i=1}^{k}a_i}{2\sum_{i=1}^{k-1}a_i}$ for all, and adjust down $a_k$ by a scale factor of $t=\frac{\sum_{i=1}^{k}a_i}{2a_k}$, and achieve a large value for LFS, and make $\sum_{i=1}{k}a_i=a_k$. Note that since $\sum_{i=1}^{k}a_i$ is not changed, and $a_1$, $\cdots$, $a_{k-1}$ are scaled up by the same factor, all conditions are still satisfied. We just need to prove that really the left hand side gets larger or equal.

Really there is just one case, but let's do three cases: 1) $k=2$, and 2) $k=n$; and 3) $2

1) $k=2$, which means $a_1

2) $k=n$. This is easy as The only value changed is $\frac{a_{n-1}}{a_n}$ but we scale up $a_{n-1}$ and down $a_n$.

3) $2a_k$ and $a_k \ge \sum_{i=1}^{k-1}a_i = 2a_{k-1}$.

So if we adjust up/down all $a_i$ and make all equalities in the condition hold, we achieve the largest value for $\frac{a_1}{a_2}+\cdots+\frac{a_{n-1}}{a_n}$, which happens to be $\frac{n}{2}$.