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# Is x^4 + y^4 > z^4? 1. x^2 + y^2 > z^2 2. x + y > z

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Manager
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Is x^4 + y^4 > z^4? 1. x^2 + y^2 > z^2 2. x + y > z [#permalink]  05 Aug 2005, 23:26
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Is x^4 + y^4 > z^4?

1. x^2 + y^2 > z^2
2. x + y > z
Director
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Re: DS-integers,,,,, [#permalink]  06 Aug 2005, 05:05
(A) for me.

In statement 2, x or y may be negative. Consider x,y,z=-10,-9,-20 and x,y,x=2,1,1. You will get different results.

In (1), the value of x^2, y^2 and z^2 must be +ve and if the sum of x^2 and y^2 is greater than z^2, then we can conclude that the same will hold true for the stem.
Manager
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But (x^2 + y^2)^2 < (x^4 + y^4)

you have forgotten the cross terms !
Manager
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Don't forget the cross terms when you square the left hand side !

Consider x=4, y=5, z=6

x + y > z
x^2 + y^2 > z^2
but
x^4 + y^4 < z^4

Consider x= 4, y=5 , z=1
x + y > z
x^2 + y^2 > z^2
and
x^4 + y^4 > z^4

Neither statement on its own or both is sufficient.
Manager
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OA is E.

I was wondering if there is a way other than plugging in numbers.
Manager
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Just look at all the cross-terms in the algebraic expansion.
Put in = instead of > to make it easier.

x + y = z
(x+y)^2 = z^2
x^2 + y^2 = z^2 - 2xy
(x^2 + y^2)^2 = (z^2 - 2xy)^2
x^4 + y^4 = z^4 + 2x^2y^2 - 4xyz^2
x^4 + y^4 = z^4 + 2x^2y^2 - 4xy(x^2+y^2)
Manager
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richardj wrote:
Just look at all the cross-terms in the algebraic expansion.
Put in = instead of > to make it easier.

x + y = z
(x+y)^2 = z^2
x^2 + y^2 = z^2 - 2xy
(x^2 + y^2)^2 = (z^2 - 2xy)^2
x^4 + y^4 = z^4 + 2x^2y^2 - 4xyz^2
x^4 + y^4 = z^4 + 2x^2y^2 - 4xy(x^2+y^2)

from that point on, how do we judge insuff from both sides (less than/greater than)? do we have to try different values?
Manager
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Hmm, good point.

Not sure I can actually finish the proof !

And you would never write all that out in the test itself.

The point is that even with x,y>0, there are terms with positive and negative signs involving powers of x and y on the right. So that makes it quite likely there are two different cases.
SVP
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Re: DS-integers,,,,, [#permalink]  09 Aug 2005, 20:54
1. x^2 + y^2 > z^2
(x^2+y^2)^2>z^4
x^4+y^4+2x^2y^2>z^4
x^4+y^4>z^4-2x^2y^2
Since x^2y^2>0, we don't know if x^4+y^4>z^4.
Insufficient

2. x + y > z
They could be different signs. Say both x and y are small positive numbers and z is a large negative. Then x^4+y^4 could be smaller than z^4. But the opposite is also possible. So insufficient.

Combined:
Not very easy to prove. The best way to go, in my opinion is by counter examples. As long as you can find a set of numbers where both 1 and 2 are satisfied to answer No to the stem then you are all set. Because it is easy to find examples to say Yes to the stem, finding the No answer would mean the answer is not unique and that 1 and 2 combined are not sufficient.
Anyways you could try with the closest numbers such as 3, 3, 4, where 3+3>4, 3^2+3^2>4^2, but 3^4+3^4<4^4.

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Re: DS-integers,,,,,   [#permalink] 09 Aug 2005, 20:54
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