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Is x^4 + y^4 > z^4 ?

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Is x^4 + y^4 > z^4 ? [#permalink] New post 20 Mar 2008, 09:19
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49% (02:12) correct 51% (00:58) wrong based on 80 sessions
Is x^4 + y^4 > z^4 ?

(1) x^2 + y^2 > z^2
(2) x+y > z

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-4-y-4-z-101358.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Oct 2013, 06:54, edited 1 time in total.
Edited the question and added the OA
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Re: DS : Is x4 + y4 > z4 [#permalink] New post 20 Mar 2008, 09:35
maverick101 wrote:
Is x^4 + y^4 > z^4

(1) X^2 + y^2 > z^2
(2) X + Y > Z


E.

1 & 2:
if x = y = 0.8 and z = 0.9 ---- yes.
if x = y = 0.7 and z = 0.9 ------- no.
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Re: DS : Is x4 + y4 > z4 [#permalink] New post 20 Mar 2008, 09:38
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E

Is x^4 + y^4 > z^4

(1) x^2 + y^2 > z^2

(2) x + y > z

It is obviously that x=10000, y=10000, z=1 satisfies both conditions and answer is "true".
So, I will try to construct example that satisfies both condition but answer is "false"
1. let x=y and z=1 for simplicity.
2. we need x<1 because x^n will decrease when the power increases.
consider the fist condition: 2*x^2>1 --> x>\sqrt{0.5} \approx 0.7
3. try x=0.8 (it satisfies both conditions): 2*0.8^4 = 2*0.64^2 < 2*0.7^2 < 2*0.49 < 1 bingo!

Here, I've tried to clarify my logic (for myself also :)) when I construct examples.....just exercise :shock:
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Re: Is x^4 + y^4 > z^4 (1) X^2 + y^2 > z^2 (2) X + Y > [#permalink] New post 10 Oct 2013, 06:50
Any alternative solution to this problem?
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Re: Is x^4 + y^4 > z^4 (1) X^2 + y^2 > z^2 (2) X + Y > [#permalink] New post 10 Oct 2013, 06:55
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fozzzy wrote:
Any alternative solution to this problem?


Is x^4+y^4>z^4?

The best way to deal with this problem is plugging numbers. Remember on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

(1) x^2+y^2>z^2
It's clear that we get YES answer very easily with big x and y (say 10 and 10), and small z (say 0).

For NO answer let's try numbers from Pythagorean triples:
x^2=3, y^2=4 and z^2=5 (x^2+y^2=7>5=z^2) --> x^4+y^4=9+16=25=z^4, so we have answer NO (x^4+y^4 is NOT more than z^4, it's equal to it).

Not sufficient.

(2) x+y>z. This one is even easier: again we can get YES answer with big x and y, and small z.

As for NO try to make z some big enough negative number: so if x=y=1 and z=-5, then x^4+y^4=1+1=2<25=z^4.

Not sufficient.

(1)+(2) As we concluded YES answer is easily achievable. For NO try the case of x^2=3, y^2=4 and z^2=5 again: x+y=\sqrt{3}+\sqrt{4}>\sqrt{5} (\sqrt{3}+2 is more than 3 and \sqrt{5} is less than 3), so statement (2) is satisfied, we know that statement (1) is also satisfied (x^2+y^2=7>5=z^2) and x^4+y^4=9+16=25=z^4. Not sufficient.

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-4-y-4-z-101358.html
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Re: Is x^4 + y^4 > z^4 (1) X^2 + y^2 > z^2 (2) X + Y >   [#permalink] 10 Oct 2013, 06:55
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