341Nice qn. a
At least two of the nonnegative real numbers $ a_{1},a_{2},...,a_{n} $ are nonzero. Decide whether a or b is larger if
$ a=\sqrt[2002]{a_{1}^{2002}+a_{2}^{2002}+\ldots+a_{n}^{2002}} $
and
$ b=\sqrt[2003]{a_{1}^{2003}+a_{2}^{2003}+\ldots+a_{n}^{2003}} $
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6 Answers
62Hint: All one needs to do is to prove the same for a1, a2 i.e. n=2
and the rest will follow..
and for that all one needs is to substitute a1=1
Why?
can someone post the complete solution now ;)
Sorry Bhargav.. couldnot prevent myself from giving this hint..
62I cant see this one hanging here unplucked :D
when ai's are zero, they dont change the inequality.. so we can assume that all the ai's are non zero..
For a1, a2 where a1≠a2 and non zero
a/a1 and b/a1 will greater or lesser exactly as a and b are.
so for n=2, the question is same as 1, a2/a1=k
so we have to see which is greater
a/a1=(1+k2002)1/2002
or
b/a1=(1+k2003)1/2003
Now someone try to finish the remaining part?
1taking k2002 common we get
a/a1= k(1+1/k2002)1/2002
using binomial approximation
a/a1=k(1+12002k2002)
similarly b/b1=k(1+12003k2003)
since 2002k2002 < 2003k2003
the result follows
ans is a is greater