good inequality !!

well its very easy ..

1/2<2/3

3/4<4/5

As n/n+1<n+1/n+2

multiplying we get n²+2n<n²+2n+1 =>0<1 which is true
similarly

999999/1000000<1000000/1000001

multiplying the above inequalities

1/2*3/4*4/5*...999999/1000000<2*4/3*..1000000/999999*(1/1000001)

let the quatity on the left side be p
then we get

p<(1/p)*(1/1000001)
thus

p<√(1/1000001)<√(1/1000000)<1/1000

@rohan 200 your solution is perfect.
i thought up another solution
\frac{n-1}{n} < (\sqrt{\frac{n-1}{n}})(\sqrt{\frac{n}{n+1}}) is true for n>1 (it is easy to check by cross multiplication)
n=2 we have
\frac{1}{2}< (\sqrt{\frac{1}{2}})(\sqrt{\frac{2}{3}})
n=4 we have

\frac{3}{4}< (\sqrt{\frac{3}{4}})(\sqrt{\frac{4}{5}})
.
.
.
n=1000000 we have
\frac{999999}{1000000}< (\sqrt{\frac{999999}{1000000}})(\sqrt{\frac{1000000}{1000001}})
multiplyng the LHS and RHS of the above inequalities we get
(\frac{1}{2})(\frac{3}{4})(\frac{5}{6}).......(\frac{999999}{1000000})<\sqrt{\frac{1}{1000001}}<\frac{1}{1000} (as successive numerators and denominators get cancelled exept last one)
and hence the inequality