let $n$ be give postive integers,and $a_{i}>0,b_{i}>0(i=1,2,\cdots,n)$.and such $$\sum_{i=1}^{n}a^2_{i}=\sum_{i=1}^{n}b^2_{i}=1$$ find the maximum of the value $$\min_{1\le i\le n}\left(\dfrac{a_{i}}{b_{i}}\right)$$

My attemp:let $$f=\max\min_{1\le i\le n}\left(\dfrac{a_{i}}{b_{i}}\right)$$ since $$a^2_{i}=1-\sum_{j\neq i}a^2_{j}$$ $$b^2_{i}=1-\sum_{j\neq i}b^2_{j}$$ so $$\left(\dfrac{a_{i}}{b_{i}}\right)^2=\dfrac{1-\sum_{j\neq i}a^2_{j}}{1-\sum_{j\neq i}b^2_{j}}$$

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    $\begingroup$ isn't this just obvious? Of course $\min(\frac{a_i}{b_i}) \le 1$, and this min can be achieved by having $a_i = b_i = \frac{1}{\sqrt{n}}$ for each $i$ $\endgroup$ Aug 30, 2019 at 15:56
  • $\begingroup$ @mathworker21 Show us, please, your obvious proof. Thank you. $\endgroup$ Aug 30, 2019 at 18:19
  • $\begingroup$ @MichaelRozenberg I did. do I really need to explain that $a_i=b_i=\frac{1}{\sqrt{n}}$ is valid and gives $\min = 1$? do I need to explain that we can't have $a_i > b_i$ for each $i$ and $\sum_i a_i^2 = \sum_i b_i^2$? what exactly do you want? $\endgroup$ Aug 30, 2019 at 19:08
  • $\begingroup$ @mathworker21 I want to see your solution. Your words are not solution. Try to post it. If you'll see down-voting it will be not mine. $\endgroup$ Aug 30, 2019 at 19:12
  • $\begingroup$ @MichaelRozenberg I'm confused. I feel like you're up to something. I think I have already said enough - I think the distinction in your mind between a non-solution and a solution is not a real one. $\endgroup$ Aug 30, 2019 at 19:32

2 Answers 2


Let $$\min_{i}\frac{a_i}{b_i}=k>0.$$

Thus, $$a_i^2\geq k^2b_i^2,$$ which gives $$\sum_{k=1}^na_i^2\geq\sum_{k=1}^nk^2b_k^2$$ or $$1\geq k^2,$$ which gives $$k\leq1.$$ The equality occurs for $$a_1=a_2=...=a_n=b_1=b_2=...=b_n=\frac{1}{\sqrt{n}},$$ which says that we got a maximal value.


We'll guess the answer is $1$.

$0=\sum_{i=1}^n \dfrac{\left(a_i^2-b_i^2\right)}{b_i}=\sum_{i=1}^n \left(a_i+b_i\right)\left(\dfrac{a_i}{b_i}-1\right)$

If there exists $k$ such that $\dfrac{a_k}{b_k}-1>0$, then $\sum_{1\le i\le n, i\ne k} \left(a_i+b_i\right)\left(\dfrac{a_i}{b_i}-1\right)=-\left(a_k+b_k\right)\left(\dfrac{a_k}{b_k}-1\right)<0 \\ \because a_i+b_i>0 \\ \therefore\text{ there must have }m\text{ that }\dfrac{a_m}{b_m}-1<0 $

Can you finish it?


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