Inequality DS

[aj5105]

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

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

(2) x+y > z


How to quickly arrive at answers for such problems?

[abhinav85]

I think IMO E

What is the OA?

[ghacker]

The thing to remember is X, Y ,Z can take any value ( it is not given that X,Y,Z are integers )

[mikeCoolBoy]

IMO E

you can solve the question by picking up numbers. I think is easy to see that if x,y and z are greater than 1 the statement is true and both statements will be sufficient
for instance x = 3, y = 4 and z = 2

we know that if a number is between 0 and 1 when you square it gets smaller so that is the kind of number you have to look for.
for z pick 1 since it does not matter how many times you square 1, the number will always be 1

statement 2)

you need to find two numbers which sum will be slightly greater than 1
let's say x=0.51, y=0.51, z =1
x + y > z ---> (0.51)^4 + (0.51)^4 > 1
I think is clear that (0.51)^4 is less than 0.5 so statement 2 is not sufficient

statement 1)

x^2 + y ^2 > z

same procedure, based on the previous example pick
x = sqrt(0.51) y = sqrt(0.51) z = 1

not sufficient


statement 1 and 2
this is the difficult part let's pick z = 1 as always and now two numbers that when square and sum will be greater than 1

x = 0.8 and Y = 0.8 z = 1

II) x + y > z
I) x^2 + y^2 > z^2 --> 0.64 + 0.64 > 1

x^4 + y^ 4 > z^4 -->2*(0.64)^2 > 1 --> 2 * 0.4 > 1

not sufficient

[rahulg83]

IMO E, I simply plugged in x^2=2, y^2=3, and z^3=4 for statement 1 and x=2,y=3,z=4 for statement 2..in both the cases the inequality in the question gets REVERSED
Remember, the tricky part was this one. In any case, you can have a huge pool of (x,y,z) that SATISFIES the above inequality..

therefore E

[aj5105]

cool answer, Rahul.