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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

$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)=\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} >\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} \\ $$ $$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)\left(\mathrm{1}+{a}_{\mathrm{3}} \right)>\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} \\ $$ $$..... \\ $$ $$.....\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)....\left(\mathrm{1}+{a}_{{n}} \right)>\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +..+{a}_{{n}} \\ $$ $${when}\:{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +...+{a}_{{n}} ={S} \\ $$ $$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)..\left(\mathrm{1}+{a}_{{n}} \right)>\mathrm{1}+{S}\:{proved} \\ $$ $${contd} \\ $$ $${let}\:{s}_{\mathrm{2}} ={a}_{\mathrm{1}} +{a}_{\mathrm{2}} \\ $$ $$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)=\mathrm{1}+{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} =\mathrm{1}+{s}_{\mathrm{2}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} \\ $$ $$\left(\mathrm{1}−{a}_{\mathrm{1}} \right)\left(\mathrm{1}−{a}_{\mathrm{2}} \right)=\mathrm{1}−\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)+{a}_{\mathrm{2}} {a}_{\mathrm{2}} =\mathrm{1}−{s}_{\mathrm{2}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} \\ $$ $$\mathrm{1}+{s}_{\mathrm{2}} >\mathrm{1}−{s}_{\mathrm{2}} \\ $$ $$ \\ $$ $${wait}\:{pls} \\ $$

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