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Question Number 55904 by gunawan last updated on 06/Mar/19
Let a_i >0, ∀_i =1, 2, 3, …2016 If (a_1 a_2 …a_(2016) )^(1/(2016)) =2 then (1+a_1 )(1+a_2 )…(1+a_(2016) )≥...
$$\mathrm{Let}\:{a}_{{i}} >\mathrm{0},\:\forall_{{i}} =\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\ldots\mathrm{2016} \\ $$$$\mathrm{If}\:\left({a}_{\mathrm{1}} {a}_{\mathrm{2}} \ldots{a}_{\mathrm{2016}} \right)^{\frac{\mathrm{1}}{\mathrm{2016}}} =\mathrm{2} \\ $$$$\mathrm{then} \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)\ldots\left(\mathrm{1}+{a}_{\mathrm{2016}} \right)\geqslant… \\ $$
Answered by mr W last updated on 06/Mar/19
for a,b>0 we have ((a+b)/2)≥(√(ab)) (1+a_1 )(1+a_2 )…(1+a_(2016) ) ≥2(√a_1 )×2(√a_2 )×...×2(√a_(2016) ) =2^(2016) (√(a_1 a_2 ...a_(2016) )) =2^(2016) (2^(2016) )^(1/2) =2^(2016) 2^(1008) =2^(3024)
$${for}\:{a},{b}>\mathrm{0}\:{we}\:{have}\:\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}} \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)\ldots\left(\mathrm{1}+{a}_{\mathrm{2016}} \right) \\ $$$$\geqslant\mathrm{2}\sqrt{{a}_{\mathrm{1}} }×\mathrm{2}\sqrt{{a}_{\mathrm{2}} }×…×\mathrm{2}\sqrt{{a}_{\mathrm{2016}} } \\ $$$$=\mathrm{2}^{\mathrm{2016}} \sqrt{{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{\mathrm{2016}} } \\ $$$$=\mathrm{2}^{\mathrm{2016}} \left(\mathrm{2}^{\mathrm{2016}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{2}^{\mathrm{2016}} \mathrm{2}^{\mathrm{1008}} \\ $$$$=\mathrm{2}^{\mathrm{3024}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19
a_1 a_2 a_3 ...a_n =2^n n=2016 ((1+a_1 )/2)≥(√(1×a_1 )) (1+a_1 )(1+a_2 )...(1+a_n )≤2^ ×(a_1 )^(1/2) ×2(a_2 )^(1/2) ..2×(a_n )^(1/n) do≥2^n (a_1 a_2 ...a_n )^(1/2) do≥2^n (2^n )^(1/2) d0≥2^(n+(n/2)) so required answer is 2^(2016+1008) (1+a_1 )(1+a_2 )...(1+a_(2016) )≥2^(3024)
$${a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} …{a}_{{n}} =\mathrm{2}^{{n}} \\ $$$${n}=\mathrm{2016} \\ $$$$\frac{\mathrm{1}+{a}_{\mathrm{1}} }{\mathrm{2}}\geqslant\sqrt{\mathrm{1}×{a}_{\mathrm{1}} }\: \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)…\left(\mathrm{1}+{a}_{{n}} \right)\leqslant\mathrm{2}^{} ×\left({a}_{\mathrm{1}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} ×\mathrm{2}\left({a}_{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} ..\mathrm{2}×\left({a}_{{n}} \right)^{\frac{\mathrm{1}}{{n}}} \\ $$$${do}\geqslant\mathrm{2}^{{n}} \left({a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${do}\geqslant\mathrm{2}^{{n}} \left(\mathrm{2}^{{n}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${d}\mathrm{0}\geqslant\mathrm{2}^{{n}+\frac{{n}}{\mathrm{2}}} \\ $$$${so}\:{required}\:{answer}\:{is}\:\mathrm{2}^{\mathrm{2016}+\mathrm{1008}} \\ $$$$\left(\mathrm{1}+{a}_{\mathrm{1}} \right)\left(\mathrm{1}+{a}_{\mathrm{2}} \right)…\left(\mathrm{1}+{a}_{\mathrm{2016}} \right)\geqslant\mathrm{2}^{\mathrm{3024}} \\ $$

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