Be a1,a2...an so a1*a2*...*a3=1
show that:
(1+a1)(1+a2)...(1+an)_>2^n
i got to
(1+a1)(1+a2)...(1+an)_<(1+(a1+a1+...+an)/n)^n
and
(1+(a1+a1+...+an)/n)^n_>2^n
but if x>y
and yz
it can be either
x>z>y
or
z>x>y
so how do i proceed?
Answers
Answered by
1
Answer:
The given expression will be minimum, when a
1
=a
2
=a
3
=...a
n
.
Now, it is given that
a
1
.a
2
.a
3
...a
n
=1
Hence, a
i
n
=1
Since a
i
is Real, therefore a
i
=1
Hence, (1+a
1
+a
1
2
)(1+a
2
+a
2
2
)...(1+a
n
+a
n
2
)
=(1+a
i
+a
i
2
)
n
=3
n
.
Similar questions