Is x^4 + y^4 > z^4?
(1) x^2 + y^2 > z^2
take x=2 and y=2 and z=2
With these numbers we have x^2 + y^2 > z^2 and x^4 + y^4 > z^4
take x=sqrt(sqrt2) and y=sqrt(sqrt2) and z=sqrt(sqrt5)
With these numbers we have
sqrt2 + sqrt2 > sqrt 5
and
2 + 2 < 5
So we found that x^2 + y^2 > z^2 and x^4 + y^4 < z^4
We have our two counter-examples.
We can tell it is insufficient.
(2) x + y > z
take x=1 and y=2 and z=-20
x+y> z and x^4 + y^4 < z^4
take x=5 and y=6 and z=1
x+y> z and x^4 + y^4 > z^4
We cannot conclude anything
Took me more than 2 minutes to find the second counter-example of 2), if someone can solve it faster.
The thing with the 1) is that
x^2 + y^2 > z^2
We square all of them
x^4 + y^4 + 2(xy)^2 > z^4 does not demonstrate anything
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