Given
1/x1+1/x2+1/x3+...+1/xn=1
for REAL numbers x1, ..., xn
and define
Y = x1+x2+x3+...+xn
If I can show that Y has it's minimum value when each the x terms is equal to n
it would follow that n^2 <= Y = 5n - 4, so 1 <= n =< 4,
So then there would only be a small number of cases to examine.
So how do we justify the assumption? We could try by induction.
The base case is n=2. Given that
1/x1+1/x2=A
we get
x1 = x2/(A^2 x2 - 1) ; Y = x1+x2 = A^2 x2^2/(A^2 x2 - 1) which is minimised for x1 = x2 = 2/A^2. In particular, with A=1, Y is minimised when x1 = x2 = 2 So the base case works.
What about the case for general n? Write
1/x1+1/x2= A = 1-1/x3-...-1/xn
then x1+x2 is minimised when x1 = x2 = 2/A^2. We can repeat this argument, putting different pairs of xs on the left hand side. This shows that Y is minimised when all the xs are equal.
So we only have to look at the cases 1 <= n <= 4, for which we get
n=1, x1 = 1
n=2 x1 = x2 = 2
n=3 x1 = 2 ; x2 = 3 ; x3 = 6
n=4 x1 = x2 = x3 = x4 = 4
Edited on May 9, 2020, 10:08 pm
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Posted by FrankM
on 2020-05-09 22:04:36 |
Full solution with explanation
