Happy new 2015 Year !!!!
What is the infimum(minimum) value of
a
2
+
a
3
+
…
+
a
2
0
1
5
a
1
+
a
1
+
a
3
+
…
+
a
2
0
1
5
a
2
+
…
+
a
1
+
a
2
+
…
+
a
2
0
1
3
+
a
2
0
1
5
a
2
0
1
4
+
a
1
+
a
2
+
…
+
a
2
0
1
3
+
a
2
0
1
4
a
2
0
1
5
?
Details and assumptions:
-
a
1
,
a
2
,
.
.
.
,
a
2
0
1
4
,
a
2
0
1
5
are positive real numbers!
- Enter your answer to two decimals!
- This problem was inspired by
"2015 is coming!!!!"
by Martin Nikolov!
The answer is 2.00.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
exactly similar approach.
but it is given that all a1,a2.......,a2105 are positive real no.
Log in to reply
Right, so his idea is to take the limit as they approach zero.
For example, consider a 1 = a 2 = 1 and a 3 = a 4 = … = a 2 0 1 5 = 0 . 0 0 0 0 0 0 0 0 0 1 .
How would you justify that you have indeed found the infimum over all possible sets of positive values?
This is not mathematically correct, but more of a reasoning way.
We want to reduce as many as possible to their minimum value, thus we can make a 3 , a 4 . . . . a 2 0 1 5 ≈ 0 (and by approx I mean VERY close to 0 like ∞ 1
Now all we have is a 2 a 1 + a 1 a 2
By AM-GM
2 a 2 a 1 + a 1 a 2 ≥ ( a 2 a 1 ) ( a 1 a 2 )
⇒ a 2 a 1 + a 1 a 2 ≥ 2