Menu Close

Suppose-a-1-a-n-are-non-negative-reals-such-that-S-a-1-a-n-lt-proof-that-1-S-1-a-1-1-a-n-1-1-s-




Question Number 36217 by Rio Mike last updated on 30/May/18
Suppose a_1 ,...,a_n ,are non−negative  reals such that S= a_1 +...+a_n <  proof that   1 + S≤ (1 + a_1 )._(...) .(1+ a_n ) ≤ (1/(1−s))
$$\mathrm{Suppose}\:{a}_{\mathrm{1}} ,…,{a}_{{n}} ,\mathrm{are}\:\mathrm{non}−\mathrm{negative} \\ $$$$\mathrm{reals}\:\mathrm{such}\:\mathrm{that}\:{S}=\:{a}_{\mathrm{1}} +…+{a}_{{n}} < \\ $$$${proof}\:{that}\: \\ $$$$\mathrm{1}\:+\:\mathrm{S}\leqslant\:\left(\mathrm{1}\:+\:{a}_{\mathrm{1}} \right)._{…} .\left(\mathrm{1}+\:{a}_{{n}} \right)\:\leqslant\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{s}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 31/May/18
(1+a_1 )(1+a_2 )=1+a_1 +a_2 +a_1 a_2 >1+a_1 +a_2   (1+a_1 )(1+a_2 )(1+a_3 )>1+a_1 +a_2 +a_3   .....  .....(1+a_1 )(1+a_2 )....(1+a_n )>1+a_1 +a_2 +..+a_n   when a_1 +a_2 +a_3 +...+a_n =S  (1+a_1 )(1+a_2 )..(1+a_n )>1+S proved  contd  let s_2 =a_1 +a_2   (1+a_1 )(1+a_2 )=1+a_1 +a_2 +a_1 a_2 =1+s_2 +a_1 a_2   (1−a_1 )(1−a_2 )=1−(a_1 +a_2 )+a_2 a_2 =1−s_2 +a_1 a_2   1+s_2 >1−s_2     wait pls
$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)=\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} >\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)\left(\mathrm{1}+{a}_{\mathrm{3}} \right)>\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \\ $$$$….. \\ $$$$…..\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)….\left(\mathrm{1}+{a}_{{n}} \right)>\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +..+{a}_{{n}} \\ $$$${when}\:{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{n}} ={S} \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)..\left(\mathrm{1}+{a}_{{n}} \right)>\mathrm{1}+{S}\:{proved} \\ $$$${contd} \\ $$$${let}\:{s}_{\mathrm{2}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)=\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} =\mathrm{1}+{s}_{\mathrm{2}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} \\ $$$$\left(\mathrm{1}−{a}_{\mathrm{1}} \right)\left(\mathrm{1}−{a}_{\mathrm{2}} \right)=\mathrm{1}−\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)+{a}_{\mathrm{2}} {a}_{\mathrm{2}} =\mathrm{1}−{s}_{\mathrm{2}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} \\ $$$$\mathrm{1}+{s}_{\mathrm{2}} >\mathrm{1}−{s}_{\mathrm{2}} \\ $$$$ \\ $$$${wait}\:{pls} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *