I got C but the answer is E.

A is insufficient because we don't know whether x and y are fractions or integers.

B is insufficient for the same reason.

so, I chose C..

Experts pls help..

Let a= x/z and b = y/z

Is a^4 + b^4 > 1

1. a^2 + b^2 >1


a^4 + b^4 + 2a^2b^2 > 1
a^4 + b^4 -2a^2b^2 > 0

a^4 + b^4 > 1/2 (insufficient)


2. a + b > 1 when z is +ve
a^2+b^2 > 1/2
a^4 + b^4 > 2a^2b^2 = 2(1/4)(1/4) = 1/8
a^4 + b^4 > 1/8 Insufficient.

Combined together, a^4 + b^4 > 1/2

Insufficient.

Forget abt z being negative.

palvarez wrote:Let a= x/z and b = y/z

Is a^4 + b^4 > 1

1. a^2 + b^2 >1


a^4 + b^4 + 2a^2b^2 > 1
a^4 + b^4 -2a^2b^2 > 0

a^4 + b^4 > 1/2 (insufficient)


2. a + b > 1 when z is +ve
a^2+b^2 > 1/2
a^4 + b^4 > 2a^2b^2 = 2(1/4)(1/4) = 1/8
a^4 + b^4 > 1/8 Insufficient.

Combined together, a^4 + b^4 > 1/2

Insufficient.

Forget abt z being negative.

Hi palvarez,
I can't get how did you come up with a^4 + b^4 + 2a^2b^2 > 1
If you don't mind pls explain it in more detail

a^2 + b^2 > 1

square on both sides.

a^4 + b^4 + 2.a^2. b^2 > 1



another obvious fact: (a2 - b^2)^2 > 0


a+b >= k, what's the min value of a^2+b^2? twice the geometric mean. Compute the minimum of the geometric mean when a = b = k/2.

Hi Palvarez,

Thanks for the response. However, how do u get this statement: -

a^4 + b^4 -2a^2b^2 > 0

Pls clarify

Thanks

LA

lance770 wrote:Hi Palvarez,

Thanks for the response. However, how do u get this statement: -

a^4 + b^4 -2a^2b^2 > 0

Pls clarify

Thanks

LA

(a^2-b^2)^2 >= 0 (a square is +ve)

a^4 +b^4 - 2a^2b^2 >= 0

a^4 + b^4 >= 2a^2b^2


Or average (arithmetic mean) >= geometric mean

I don't understand the explanation given by Palvarez... No offence.

Look at this thread for explanations given by Ron Purewal and Ian Stewart.

https://www.beatthegmat.com/gmatprep-is- ... 23339.html

Just adding more steps for others to understand


\Assume a = x/z, b = y/z



Is a^4 + b^4 > 1

1. a^2 + b^2 >1

(a^2+b^2)^2 > 1, squaring a positive doesn't change the inequality.

a^4 + b^4 + 2a^2b^2 > 1 ---(1)

(a^2-b^2) >= 0 (any square is +ve; this is the same used in deriving AM >= GM inqualitty, gm = geometric mean and am = arithmetic mean)

a^4 + b^4 -2a^2b^2 >= 0 -- (2)

Add (1) and (2): a^4 + b^4 >= 1/2 (insufficient)


2. a + b > 1 when z is +ve

a^2+b^2 + 2ab > 1 (square the above) --(3)

(a-b)^ 2>= 0

a^2 + b^2 -2ab >= 0 --- (4)

Add (3) and (4): a^2+b^2 > 1/2

Now square it again and use (a^2 - b^2)^2 and sum them up, you will get a^4+b^4 >= 1/8 (insufficient)


------------------------------------
There is a faster way to do all these things

a^2+b^2 >= 2ab (arithmetic mean >= geometric mean)

Now we need to get the minimum value of 2ab when a + b > 1: how to go about?

just assume a = b and solve a+b =1, ya end up with a = b = 1/2
Substitute these in the above inequality.

2ab = 2(1/2)(1/2) = 1/2

a^2 + b^2 > =1/2

a^4+b^4 >= 2a^2b^2 (arith mean >= geometric mean)


What is a^2 when a^2 = b^2 and a^2 + b^2 = 1/2
a^2 = b^2 = 1/4

therfore a^4 + b^4 >= 2(1/4(1/4)

a^4+b^4 >= 1/8

1. X^2 + y^2 > z^2 => x^4 + y^4+ 2x^2y^2 > z^4

2. x+y > z => x^2 + y^2 + 2xy > z^2

using 1 and 2 : z^2 + 2xy > z^2 => 2xy>0 we can only infer x and y are of same sign

Answer is E.

GmatVerbal,

I love how you can solve this in 3 lines where others take half a page. But your concise solution fails to explains your rationale. You may understand it, but I sure don't. So, if your intent is for others to wonder in amazement about how it is that you can solve difficult problems in 3 lines -

Oooh, aaaaah...