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 $$\left(\dfrac{a_{i}}{b_{i}}\right)^2=\dfrac{1-\sum_{j\neq i}a^2_{j}}{1-\sum_{j\neq i}b^2_{j}}$$
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?
