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Question Number 211605 by MrGaster last updated on 14/Sep/24
              Let a_1 ,a_2 ,…a_n                 Is n real numbers.              All fall in the interval (−1,1)  ________________________  (1)Prove that:                      Π_(1≤i,j≤n) ((1+a_i a_j )/(1−a_i a_j ))≥1  (2) Determine the necessary   andsufficient conditions forl  equaity in the inequality.
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Let}\:\boldsymbol{{a}}_{\mathrm{1}} ,\boldsymbol{{a}}_{\mathrm{2}} ,\ldots\boldsymbol{{a}}_{\boldsymbol{{n}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Is}\:\mathrm{n}\:\mathrm{real}\:\mathrm{numbers}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{All}\:\mathrm{fall}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\left(\mathrm{1}\right)\mathrm{Prove}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{1}\leq\boldsymbol{{i}},\boldsymbol{{j}}\leq\boldsymbol{{n}}} {\prod}\frac{\mathrm{1}+\boldsymbol{{a}}_{\boldsymbol{{i}}} \boldsymbol{{a}}_{\boldsymbol{{j}}} }{\mathrm{1}−\boldsymbol{{a}}_{\boldsymbol{{i}}} \boldsymbol{{a}}_{\boldsymbol{{j}}} }\geq\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{necessary}\: \\ $$$$\mathrm{andsufficient}\:\mathrm{conditions}\:\mathrm{forl} \\ $$$$\mathrm{equaity}\:\mathrm{in}\:\mathrm{the}\:\mathrm{inequality}. \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 15/Sep/24
   mr   hardmath?
$$\:\:\:{mr}\:\:\:{hardmath}? \\ $$
Answered by a.lgnaoui last updated on 14/Sep/24
E=1 ⇒ 2Πa_i a_j =0        So  ;if one of a_i or a_j =0  then Π ((1+a_i a_j )/(1−a_i a_j ))=1
$$\mathrm{E}=\mathrm{1}\:\Rightarrow\:\mathrm{2}\Pi\mathrm{a}_{\mathrm{i}} \mathrm{a}_{\mathrm{j}} =\mathrm{0}\:\:\:\:\:\: \\ $$$$\mathrm{So}\:\:;\mathrm{if}\:\mathrm{one}\:\mathrm{of}\:\mathrm{a}_{\mathrm{i}} \mathrm{or}\:\mathrm{a}_{\mathrm{j}} =\mathrm{0}\:\:\mathrm{then}\:\Pi\:\frac{\mathrm{1}+\mathrm{a}_{\mathrm{i}} \mathrm{a}_{\mathrm{j}} }{\mathrm{1}−\mathrm{a}_{\mathrm{i}} \mathrm{a}_{\mathrm{j}} }=\mathrm{1} \\ $$$$ \\ $$
Commented by mr W last updated on 15/Sep/24
a_i  or a_j  is not a particular number!  for example you can not say:   such that Σ_(i=1) ^n a_i =0,  ⇒a_i =0.
$${a}_{{i}} \:{or}\:{a}_{{j}} \:{is}\:{not}\:{a}\:{particular}\:{number}! \\ $$$${for}\:{example}\:{you}\:{can}\:{not}\:{say}:\: \\ $$$${such}\:{that}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} =\mathrm{0}, \\ $$$$\Rightarrow{a}_{{i}} =\mathrm{0}. \\ $$

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