1) x^2 + y^2 > Z^2 X^2, Y^2, Z^2 are always positive : DS Archive
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# 1) x^2 + y^2 > Z^2 X^2, Y^2, Z^2 are always positive

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1) x^2 + y^2 > Z^2 X^2, Y^2, Z^2 are always positive [#permalink]

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23 Jul 2009, 19:29
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1) x^2 + y^2 > Z^2

X^2, Y^2, Z^2 are always positive numbers, so squaring on both sides does not have any effect on the inequality
i.e
(x^2 + y^2)^2 > (Z^2)^2
X^4 + Y^4 + 2X^2Y^2 > Z^4

All the values mentioned above are positive numbers. So obviously
X^4 + Y^4 > Z^4

2) X + Y > Z

In this scenario nothing is mentioned about X, Y & Z. They might be either positive or negative numbers.

Lets suppose X= 1, Y = 1, Z = -50

2> -50

squaring both sides, 4< 2500. Squaring again will retain the same inequality as both the sides have positive numbers

X^4 + Y^4 < Z^4

but if X, Y, Z are +ve then X^4 + Y^4 > Z^4 holds true.

So 1 alone is sufficient but 2 is not required and the answer is "A".
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Re: DS Inequalities Problem for Math Wiz [#permalink]

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23 Jul 2009, 22:08
Aleehsgonji wrote:
1) x^2 + y^2 > Z^2

X^2, Y^2, Z^2 are always positive numbers, so squaring on both sides does not have any effect on the inequality
i.e
(x^2 + y^2)^2 > (Z^2)^2
X^4 + Y^4 + 2X^2Y^2 > Z^4

All the values mentioned above are positive numbers. So obviously
X^4 + Y^4 > Z^4

2) X + Y > Z

In this scenario nothing is mentioned about X, Y & Z. They might be either positive or negative numbers.

Lets suppose X= 1, Y = 1, Z = -50

2> -50

squaring both sides, 4< 2500. Squaring again will retain the same inequality as both the sides have positive numbers

X^4 + Y^4 < Z^4

but if X, Y, Z are +ve then X^4 + Y^4 > Z^4 holds true.

So 1 alone is sufficient but 2 is not required and the answer is "A".

No offence but neither is the approach to this problem correct nor is the solution.

Lets start with the first statement:

I. x^2+y^2>z^2
Squaring both sides we get:
x^4+y^4+2x^2y^2 > z^4
Since 2x^2y^2 is a positive quantity, on it's own this expression is insufficient to determine whether x^4+y^4>z^4

Alternatively,
Let's try and plug-in certain values in this expression:
a. x=1; y=0; z=0
x^2+y^2>z^2
Clearly, x^4+y^4>z^4 holds true here since [((0+1)^2)^2]>[(0)^2]^2
TRUE

b. x=1; y=3; z=2
z^2=4
x^2+y^2=(1+9)=10
Clearly x^2+y^2>z^2

x^4=1; y^4=81; z^4=16
x^4+y^4 > z^4
TRUE

c. x=4; y=5; z=6
z^2=36
x^2+y^2=(16+25)=41
Clearly x^2+y^2>z^2

x^4=256; y^4=625; z^4=1296
x^4+y^4 < z^4
FALSE

Hence, Statement I is INSUFFICIENT

Let's look at the second statement:
II. x+y>z

Squaring both sides
(x+y)^2>z^2
x^2+y^2+2xy > z^2

Squaring again we get:
x^4+4x^3y+6x^2y^2+4xy^3+y^4 > z^4

This expression is Insufficient on it's own to determine whether x^4+y^4 > z^4 since don't know whether x,y,z are positive or negative integers.

Alternatively,
Lets plug-in certain values for x,y and z:
a. x=1; y=1; z=0
(1^4 + 1^4) > 0^4
Hence, x^4+y^4 > z^4
TRUE

b. x=2; y=3; z=4
(2^4+3^4)= (16+81)=97
4^4=256
x^4+y^4 < z^4
FALSE

c. x=1; y=1; z=-10
(x^4+y^4)=2
z^4= 10000
x^4+y^4 < z^4
FALSE

Hence, Statement II is INSUFFICIENT

Combining both the statements doesn't help either as there is no common ground between the two statements which could lead to a solution to the question.

Hence the Answer is E
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Re: DS Inequalities Problem for Math Wiz [#permalink]

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24 Jul 2009, 01:19
x=3, y=3, z=4
x+y>z
x^2+y^2>z^2
But
x^4+y^4<z^4

x=10, y=10, z=1
x+y>z
x^2+y^2>z^2
And
x^4+y^4>z^4

So E
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Re: DS Inequalities Problem for Math Wiz [#permalink]

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24 Jul 2009, 01:27
christinee wrote:
Please solve:

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

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

2) x+y>z

2) x+y>z
we have: x^2+y^2>= [(x+y)^2] . 1/2 > z^2
So in this problem (1) is relatively stronger than (2). So if statement 1 is not suff, st 2 is too.
Statement 1: x^2 + y^2 > z^2
x^4+y^4 >= [(x^2+y^2)^2] . 1/2 > z^4 . 1/2
So statement 1 just determines that x^4 + y^4 > z^4 . 1/2 so insuf
E for me.
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Re: DS Inequalities Problem for Math Wiz [#permalink]

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24 Jul 2009, 04:33
christinee wrote:
Please solve:

Is x^4 + y^4 > z^4? rephrase is x^4+y^4-z^4>0

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

2) x+y>z

from 1 ( square)

x^4+2x^2*y^2+y^4 > z^4

x^4+y^4 -z^4> 2x^2*y^2 thus the question became (WHETHER 2x^2*y^2 >0 we dont have enough info as x or y could = 0.......insuff

from 2

x+y>z......insuff
both
still insuff.....E
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Re: DS Inequalities Problem for Math Wiz [#permalink]

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24 Jul 2009, 05:17
If you want to choose numbers here, you might notice that S1 looks a lot like the Pythagorean Theorem. In any right triangle with hypotenuse c, and legs a and b, we always have a + b > c, but a^2 + b^2 is not greater than c^2 (they're equal). So if we make x^2, y^2 and z^2 the lengths of three sides of a right triangle, we'll get numbers that demonstrate that x^4 + y^4 does not need to be larger than z^4. If you let, say, x^2 = 3, y^2 = 4 and z^2 = 5 (so x = root(3), y root(4) = 2, and z = root(5)), then x+y > z, x^2 + y^2 > z^2, but x^4 + y^4 = z^4.
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Re: DS Inequalities Problem for Math Wiz [#permalink]

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24 Jul 2009, 10:12
IanStewart wrote:
If you want to choose numbers here, you might notice that S1 looks a lot like the Pythagorean Theorem. In any right triangle with hypotenuse c, and legs a and b, we always have a + b > c, but a^2 + b^2 is not greater than c^2 (they're equal). So if we make x^2, y^2 and z^2 the lengths of three sides of a right triangle, we'll get numbers that demonstrate that x^4 + y^4 does not need to be larger than z^4. If you let, say, x^2 = 3, y^2 = 4 and z^2 = 5 (so x = root(3), y root(4) = 2, and z = root(5)), then x+y > z, x^2 + y^2 > z^2, but x^4 + y^4 = z^4.

Thanks Ian..Its all about identifying the pattern
Re: DS Inequalities Problem for Math Wiz   [#permalink] 24 Jul 2009, 10:12
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# 1) x^2 + y^2 > Z^2 X^2, Y^2, Z^2 are always positive

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