For statement 1, start off by raising everything to the second power:

x^2 + y^2 > z^2 --- raise to second power to get:

x^4 + y^4 + 2(x^2)(y^2) > z^4 --- you can't algebraically demonstrate that x^4 + y^4 > z^4, since you've got that 2(x^2)(y^2) over there. That amount is a positive one (since it's the product of three positive numbers), so you don't really know if x^4 + y^4 is indeed greater than z^4 or if it needs the 2(x^2)(y^2) to make it bigger.

I actually tried going against my better judgement and tried to number pick my way through it, but it proved much harder than I thought. But here are two examples:
a. x = 4, y = 0 and z = 2. In this case, we do have that x^2 + y^2 > z^2 and also x^4 + y^4 > z^4.

b. This example was harder to build. I initially thought of fractions, but I realized that it was a dead end. For z, you need something that gets bigger faster after the second power. I eventually picked x = 5, y = 4 and z = 6.

x^2 = 25
y^2 = 16
z^2 = 36
We of course respect the fact that x^2 + y^2 > z^2, since 41 > 36.

BUT:

x^4 = 625
y^4 = 256
z^4 = 216*6, which is greater than 1000.
In this case, you get that x^4 + y^4 < z^4, since x^4 + y^4 < 1000, while z^4 > 1000.

As you can see, this strategy of picking numbers takes two things:
- good knowledge of powers
- patience

This is why for me at least the algebraic method is a lot more comfortable.

Thanks.

If we combine A and B....then also answer is E...

Can you please tell how to make it E.....using both A and B. And without using putting numbers.....as putting numbers is a very tedious and risk prone method.

DanaJ wrote:For statement 1, start off by raising everything to the second power:

x^2 + y^2 > z^2 --- raise to second power to get:

x^4 + y^4 + 2(x^2)(y^2) > z^4 --- you can't algebraically demonstrate that x^4 + y^4 > z^4, since you've got that 2(x^2)(y^2) over there. That amount is a positive one (since it's the product of three positive numbers), so you don't really know if x^4 + y^4 is indeed greater than z^4 or if it needs the 2(x^2)(y^2) to make it bigger.

Thanks DanaJ. But for the case of insufficiency ..... we have to prove that atleast one case <= and atleast one case is >.

Can you please help me prove the same.....without putting the values.....because putting values is very tricky and very error prone......by hit and trial I was able to come up with one set of values....i.e. x, y = 3,3......but want to learn the general approach to solve this problem.....instead of relying on hit and trial approach....which takes a huge amount of time.