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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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    $\begingroup$ 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 $\frac{a_2}{a_3}, \frac{a_3}{a_4},\ldots$ ? $\endgroup$ – quasi Jan 7 '17 at 18:36
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    $\begingroup$ How do you get $a_k+a_k \leq a_{k+1}$ from (4) and (5)? $\endgroup$ – rtybase Jan 7 '17 at 18:42
  • $\begingroup$ Oops -- I fell in the same trap that yasir fell in -- ignore my first comment. Instead, look at rtybase's objection. $\endgroup$ – quasi Jan 7 '17 at 18:46
  • $\begingroup$ @rtybase Subtracting (5) from (4) $\endgroup$ – MattG88 Jan 7 '17 at 18:49
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    $\begingroup$ @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). $\endgroup$ – quasi Jan 7 '17 at 19:13
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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) $2<k<n$.

1) $k=2$, which means $a_1<a_2$. Now The only values changed are $\frac{a_{1}}{a_2}$ and $\frac{a_{2}}{a_{3}}$, and we want to show that $\frac{a_{1}}{a_2} + \frac{a_{2}}{a_{3}} \le \frac{sa_{1}}{ta_2} + \frac{ta_{2}}{a_{3}}$. This is easy as $s=\frac{a_1+a_2}{2a_1}$ and $t=\frac{a_1+a_2}{2a_2}$, and $sa_1=ta_2=\frac{a_1+a_2}{2}$: $$\frac{a_{1}}{a_2} + \frac{a_{2}}{a_{3}} \le \frac{a_{1}}{a_1} + \frac{a_{1}}{a_{3}} = \frac{sa_{1}}{ta_2} + \frac{a_{1}}{a_{3}} < \frac{sa_{1}}{ta_2} + \frac{ta_{2}}{a_{3}}, $$ where the first inequality requires a simple check.

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) $2<k<n$. Now The only values changed are $\frac{a_{k-1}}{a_k}$ and $\frac{a_{k}}{a_{k+1}}$, and we want to show that $\frac{a_{k-1}}{a_k} + \frac{a_{k}}{a_{k+1}} \le \frac{sa_{k-1}}{ta_k} + \frac{ta_{k}}{a_{k+1}}$. Note that since the equality holds for $k-1$, we have $2a_{k-1}=\sum_{i=1}{k-1}a_i$. So $sa_{k-1}= \frac{2a_{k-1}+a_k}{4}$ and $ta_k=\frac{2a_{k-1}+a_k}{2}=2sa_{k-1}$. So we want to prove that $$\frac{a_{k-1}}{a_k} + \frac{a_{k}}{a_{k+1}} \le \frac{1}{2} + \frac{2a_{k-1}+a_k}{2a_{k+1}}, $$ which is equivalent to the following after multiplying $2a_ka_{k+1}$: $$2a_{k-1}a_{k+1}+2a_k^2 \le a_ka_{k+1}+(2a_{k-1}+a_k)a_k, $$ which is equivalent to $$(a_{k+1}-a_k)(a_k-2a_{k-1}) \ge 0,$$ which is true since $a_{k+1}>a_k$ 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 $\frac{n}{2}$.

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