Question

Be a1,a2...an so a1*a2*...*an=1



show that:

(1+a1)(1+a2)...(1+an)_>2^n


I got to

(1+a1)(1+a2)...(1+an)_ 2^n


but if x>y

and y z

it can be either

x>z>y

or

z>x>y


so how do i proceed?

Answer 1

Answer:

Be a1,a2...an so a1*a2*...*an=1

Be a1,a2...an so a1*a2*...*an=1show that:

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^n

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^nI got to

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^nI got to(1+a1)(1+a2)...(1+an)_ 2^n

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^nI got to(1+a1)(1+a2)...(1+an)_ 2^nbut if x>y

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^nI got to(1+a1)(1+a2)...(1+an)_ 2^nbut if x>yand y z

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^nI got to(1+a1)(1+a2)...(1+an)_ 2^nbut if x>yand y zit can be either

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^nI got to(1+a1)(1+a2)...(1+an)_ 2^nbut if x>yand y zit can be eitherx>z>y

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^nI got to(1+a1)(1+a2)...(1+an)_ 2^nbut if x>yand y zit can be eitherx>z>yor

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^nI got to(1+a1)(1+a2)...(1+an)_ 2^nbut if x>yand y zit can be eitherx>z>yorz>x>y

Be a1,a2...an so a1*a2*...*an=1show that:(1+a1)(1+a2)...(1+an)_>2^nI got to(1+a1)(1+a2)...(1+an)_ 2^nbut if x>yand y zit can be eitherx>z>yorz>x>yso how do i proceed?