Question Number 162995 by HongKing last updated on 02/Jan/22
$$\mathrm{if}\:\:\mathrm{a}_{\boldsymbol{\mathrm{k}}} \:>\:\mathrm{0}\:\:;\:\:\mathrm{k}\:=\:\overline {\mathrm{1},\mathrm{5}} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{exists}\:\:\boldsymbol{\mathrm{i}},\boldsymbol{\mathrm{j}}\in\overline {\mathrm{1},\mathrm{5}}\:\:\mathrm{such}\:\mathrm{that}: \\ $$$$\mathrm{0}\:\leqslant\:\frac{\mathrm{a}_{\boldsymbol{\mathrm{j}}} \:-\:\mathrm{a}_{\boldsymbol{\mathrm{i}}} }{\mathrm{1}\:+\:\mathrm{a}_{\boldsymbol{\mathrm{i}}} \mathrm{a}_{\boldsymbol{\mathrm{j}}} }\:\leqslant\:\sqrt{\mathrm{2}}\:-\:\mathrm{1} \\ $$
Answered by mindispower last updated on 03/Jan/22
![a_k =tg(β_k ),β_k ∈]0,(π/2)[,x→tg(x) bijection [0,(π/2)[→[0,∞[ α_k −a_i =((tg(β_k )−tg(β_i ))/(1+tg(β_k )tg(β_i )))=tg(β_k −β_i ) ⇔to show∃i,j such 0≤tg(β_j −β_i )≤(√2)−1 [0,(π/2)[]=∪_(k=0) ^3 [((kπ)/8),((k+1)/2)π[...(E) if ∃ i,j such β_i =β_j true tg(β_i −β_j )=0∈[0,(√2)−1[ suppose ∀(i,j)∈[1,5] β_i #β_j ⇒∃(i,j),∃k∈[0,] such That β_i ,β_(j ) ∈[((kπ)/8),((k+1)/8)π[ 5 number withe 4 interval ” 0≤∣β_i −β_j ∣≤(π/8) 0≤tg∣(β_j −β_i )∣≤tg((π/8)) tg((π/8))=((2sin^2 ((π/8)))/(sin(2.(π/8))))=((1−cos((π/4)))/(sin((π/4))))=(√2)−1 ⇒∃(i,j) such that0≤ ((tg(β_i )−tg(β_j ))/(1+tg(β_i )tg(β_j )))=((a_i −a_j )/(1+a_i a_j ))≤(√2)−1](https://www.tinkutara.com/question/Q163063.png)
$$\left.{a}_{{k}} ={tg}\left(\beta_{{k}} \right),\beta_{{k}} \in\right]\mathrm{0},\frac{\pi}{\mathrm{2}}\left[,{x}\rightarrow{tg}\left({x}\right)\:{bijection}\:\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\rightarrow\left[\mathrm{0},\infty\left[\right.\right.\right.\right.\right. \\ $$$$\alpha_{{k}} −{a}_{{i}} =\frac{{tg}\left(\beta_{{k}} \right)−{tg}\left(\beta_{{i}} \right)}{\mathrm{1}+{tg}\left(\beta_{{k}} \right){tg}\left(\beta_{{i}} \right)}={tg}\left(\beta_{{k}} −\beta_{{i}} \right) \\ $$$$\Leftrightarrow{to}\:{show}\exists{i},{j}\:{such} \\ $$$$\mathrm{0}\leqslant{tg}\left(\beta_{{j}} −\beta_{{i}} \right)\leqslant\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\right]=\underset{{k}=\mathrm{0}} {\overset{\mathrm{3}} {\cup}}\left[\frac{{k}\pi}{\mathrm{8}},\frac{{k}+\mathrm{1}}{\mathrm{2}}\pi\left[…\left({E}\right)\right.\right.\right. \\ $$$${if}\:\exists\:{i},{j}\:{such}\:\beta_{{i}} =\beta_{{j}} \:{true}\:{tg}\left(\beta_{{i}} −\beta_{{j}} \right)=\mathrm{0}\in\left[\mathrm{0},\sqrt{\mathrm{2}}−\mathrm{1}\left[\right.\right. \\ $$$${suppose}\:\forall\left({i},{j}\right)\in\left[\mathrm{1},\mathrm{5}\right]\:\beta_{{i}} #\beta_{{j}} \\ $$$$\Rightarrow\exists\left({i},{j}\right),\exists{k}\in\left[\mathrm{0},\right]\:{such}\:{That}\:\beta_{{i}} ,\beta_{{j}\:} \in\left[\frac{{k}\pi}{\mathrm{8}},\frac{{k}+\mathrm{1}}{\mathrm{8}}\pi\left[\right.\right. \\ $$$$\mathrm{5}\:{number}\:{withe}\:\mathrm{4}\:{interval}\:'' \\ $$$$\mathrm{0}\leqslant\mid\beta_{{i}} −\beta_{{j}} \mid\leqslant\frac{\pi}{\mathrm{8}} \\ $$$$\mathrm{0}\leqslant{tg}\mid\left(\beta_{{j}} −\beta_{{i}} \right)\mid\leqslant{tg}\left(\frac{\pi}{\mathrm{8}}\right) \\ $$$${tg}\left(\frac{\pi}{\mathrm{8}}\right)=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}{{sin}\left(\mathrm{2}.\frac{\pi}{\mathrm{8}}\right)}=\frac{\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{4}}\right)}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\Rightarrow\exists\left({i},{j}\right)\:{such}\:{that}\mathrm{0}\leqslant\:\frac{{tg}\left(\beta_{{i}} \right)−{tg}\left(\beta_{{j}} \right)}{\mathrm{1}+{tg}\left(\beta_{{i}} \right){tg}\left(\beta_{{j}} \right)}=\frac{{a}_{{i}} −{a}_{{j}} }{\mathrm{1}+{a}_{{i}} {a}_{{j}} }\leqslant\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by HongKing last updated on 03/Jan/22
$$\mathrm{Perfect}\:\mathrm{solution}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{than}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$