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

[zchampz]

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

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

How can we solve this?

Thanks,
Champ

[agha79]

What is the OA for this question? I am getting "E" for an answer here.
Starting with statement II:
If X and Y are both fractions greater than ½ and Z is "1" than when we raise X and Y to 4 it will reduce its value and Y would stay the same.
Now with Statement I:
I couldn’t come of with numbers but I kind of used the same logic. If all are fractions but are close fraction it is possible that when we take a square X and Y are greater than Z but when we raise the Power to 4 Z is larger

[zchampz]

Answer is E.

I tried to plug-in the integers and fractions ..

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

If we plug-in integers ----> x^2=1, y^2=4 and z^2=16 then
x^2 + y^2 > z ^2 => 1 + 4 > 16 => 5 > 16
In this case, x^4 + y^4 > z ^4 => 1 + 16 > 256 => 17 > 256 --->TRUE

However, if x^2=0.2, y^2=0.3 and z^2=0.6 then
x^2 + y^2 > z ^2 => 0.2+ 0.3 > 0.6 => 0.5 > 0.6
In this case, x^4 + y^4 > z ^4 => 0.04 + 0.09 > 0.36 => 0.13 > 0.36---> FALSE
Statement 1 is not sufficient

2) X + y > z
If we plug-in integers ----> x=1, y=2 and z=4 then
x^4 + y^4 > z ^4
=> 1 + 16 > 256 => 17 > 256 --->TRUE

However, if x=0.2, y=0.3 and z=0.6 then
In this case, x^4 + y^4 > z ^4 => 0.0016 + 0.0081 > 0.1296 => 0.0097 > 0.1296---> FALSE
Statement 2 is not sufficient

(1) & (2) both together are also not sufficient.


Any better way to solve this? I guess it is more conceptual than an approach.

[esledge]

I'm not sure there is any better approach than picking numbers. There are several clues:

a. The algebraic expressions (x+y), (x^2 + y^2), and (x^4+y^4) don't relate to each other via the common operations, such as squaring or multiplying one by another. Things might be different if we had (x^2-y^2), (x-y), and (x+y), for instance. Don't do algebra when there's no productive algebra to do.

b. The added squared terms in the question and (1) limit us to nonnegative values, reducing this question to 0, pos fraction, and pos integer cases.

c. We can suspect insufficiency because of reason (a.) above. Proving sufficiency by picking numbers is a terrible idea (how do you know when you are "done"?), but proving insufficiency by picking numbers is GREAT. All you need is a Yes example and a NO example (or two different value results on a value question).

[zchampz]

Thanks for clues.

[umeshkathuria]

One or 2 alone are not suff.
Lets consider option D or E
from 2 (x+y)^2>z^2
=> x^2+y^2+2xy
(x+y)^4 = x^4+y^4+[4x^2*y^2+2(x^2*y^2+2x.y^3+2yx^3)]
we don't know anything about the sign of the terms in the [parenthesis]...thus ans should be E

[akhp77]

Statement 1:
x^2 + y^2 > 0
z^2 > 0

x^2 + y^2 > z^2
squire both sides

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

2 * x^2 * y^2 can take any +ve value but we do not know how much.

x^4 + y^4 >, =, or < z^4

Insufficient

Statement 2:
x, y and z may be +ve or -ve
Lets assume all +ve and try to attack on this inequality

x + y > z
squire both sides

x^2 + y^2 + 2xy > z^2
2xy may take any value as explained above
x^2 + y^2 >, =, or < z^2

and hence x^4 + y^4 >, =, or < z^4

Insufficient

Statement 1 and 2:
Here we have same problem x^4 + y^4 >, =, or < z^4
We can't get the result.

Insufficient

Ans E

[StaceyKoprince]

good work!

[saptadeepc]

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

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


If we simplify the question :-

(x^2 + y^2)^2 - 2*x^2*y^2 > z^4

since -- 2*x^2*y^2 is always positive, we can say that

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

or

|x^2 + y^2| > |z^2|

since they are all positive

x^2 + y^2 > z^2

which is the statement I. Please correct the flaw in my approach !

[RonPurewal]

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

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


If we simplify the question :-

(x^2 + y^2)^2 - 2*x^2*y^2 > z^4

since -- 2*x^2*y^2 is always positive, we can say that

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

or

|x^2 + y^2| > |z^2|

since they are all positive

x^2 + y^2 > z^2

which is the statement I. Please correct the flaw in my approach !

the problem is not a flaw IN your approach -- the problem is the approach itself: the entire approach is backward.
you are assuming that the question is a known fact -- and treating the known fact (statement 1) as a question!

this is problematic, for reasons that should be obvious. for instance, if you have:
question: Is x > 2?
statement (1): x > 0
... then statement (1) is insufficient, but, using your approach here, you would actually conclude that it is sufficient!

[namnam123]

very hard. I can not do. but I find out the solution.

froM both 1 and 2, the answer YES is easy to prove by picking number x >y>z>1

now WE NEED TO PROVE THAT ANSWER CAN BE NO FOR BOTH 1 and 2

we need to pick numbers so that WHEN WE EXPONENT, WE NEED TO INCREASE or KEEP THE SAME Z^4 AND DECREASE X^4 + Y^4 AND KEEP THE SUM OF X^2 AND Y^2 >Z^2. we can do this by choose x<1, y<1 and Z>1

for simplicity, choose z=1 to keep the right part of inequality the same.

when z=1, the right part of inequality is the same
choose x, y so that x+y>1 and x^2+y^2>1 but x^4+y^4<1=z

x=y=0.8 is ok, x=y=0.7 is not ok.

VERY HARD.

[RonPurewal]

i'm going to lock this thread. please read the following link, and post any further questions on it:
post18074.html#p18074

thanks.