easy way required to solve this

[divya23]

please explain the easiest way to solve this....
is x^4+y^4>z^4

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


ans = none

[vikram4689]

please post the full equation, inequalities cannot be solved unless limits on the variables are not known

[divya23]

this is the full question nothing else was given

[Frankenstein]

Hi,
From(1): x^2+y^2>z^2
if x=3, y=3, z=4 x^4+y^4 = 162 < z^4(256)
if x=2, y=2, z= 1 x^4+y^4>z^4
Not sufficient

From(2):
if x=3, y=3, z=4 x^4+y^4 = 162 < z^4(256)
if x=2, y=2, z= 1 x^4+y^4>z^4
Not sufficient

Both (1)&(2): Use the same set of values as those used in (1)&(2)
Not sufficient

Hence, E

[vikram4689]

this is the full question nothing else was given

Well in that case i think answer would always be E because even if we are able to get some consistency bu putting values, we can always reverse values of x,y & z to get the opposite result.

[amit2k9]

observe that the exponents here are even.Hence we need not try negative values.

checking for 1,1,1 and 1,2,2.

a for 1,1,1 LHS=RHS. for 1,2,2 LHS > RHS. not sufficient.

b same as above. not sufficient.

a+b not sufficient.

E it is.

[Frankenstein]

a for 1,1,1 LHS=RHS. for 1,2,2 LHS > RHS. not sufficient.

For 1,1,1 LHS(2) > RHS(1).