SEe rezolva prin inductie
Fie a1 ,a2 >0 a .i a1*a2=1=> a1=1/a2
(1+a1)*(1+a2)=(1+1/a2)*(1+a2)=1+a2+1/a2+a2/2=1+1+(a2+1/a2)=2+(a2+1/2)
Dar a2+1/a2>2 =>2+(a2+1/a2)>2²
Presupunem Pn adevarata adica a1*a2*...an=1 =>P(n+1) adevarata.adica (a1*a2...*an)*an+1=1 <=> 1*an+1=1=> an+1=1
Deci
Pn=(1+a1)*...(1+an)>2^n
Pn+1: (1+a1)*...(1+an)*(1+an+1)>2^(n+1)
2^n*(1+an+1)>2^(n+1) <=>
2^n*(1+1)>2^(n+1)
2^n*2>2^(n+1) evident
Pn=>Pn+1