Inspired by Daniel Liu
Consider all sets of real numbers x , y , z which satisfy the equation
x y 2 + y z 2 + z x 2 = 3
Let the infimum value of z 2 x 2 + y 2 + x 2 y 2 + z 2 + y 2 z 2 + x 2 be denoted as n .
For how many sets of ordered triples ( x , y , z ) which satisfy the initial equation, also satisfy
z 2 x 2 + y 2 + x 2 y 2 + z 2 + y 2 z 2 + x 2 = n ?
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1 solution
I think it's kinda embarrassing that I accidentally got this question wrong.
Shame on me.
Also why did you choose to use "infimum" instead of "minimum"? Or is "infinium" a typo?
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Tsk tsk :) It's interesting in part because there are 2 cases, which lead to 1+3 equality cases.
I intentionally chose "infimum". It is worthwhile to get used to this term, and avoid a lot of confusion that results otherwise.
2 different equality 1. x=y=z 2. |x|=y=z in this there is three combination Please correct me if I'm wrong.
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Right, there are 2 different equality cases like you stated.
In the first case, there is 1 solution.
In the second case, there are 3 solutions, corresponding to
x
=
y
=
−
z
,
x
=
−
y
=
z
,
−
x
=
y
=
z
.
Hence, there are a total of 4 solutions.
This problem is inspired by Daniel Liu's problem Squared Inequality .
I made the observation that few people would notice that there are 2 different equality cases, which yield a total of 4 solutions.