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

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

OA : E

How to solve this w/o plugging numbers ..Any equation types ?

number plugging is the best way to go.

you can try squaring statement (1), yielding x^4 + 2(x^2)(y^2) + y^4 > z^4, but that's inconclusive for this question because 2(x^2)(y^2) is a positive number that could swing the inequality either way. in other words, if you take that term away, then the resulting left-hand expression (x^4 + y^4) could potentially be greater or less than z^4, depending on the size of the 2(x^2)(y^2) that was taken away. (notice that if you KNOW x^4 + y^4 > z^4, then you know conclusively that x^2 + y^2 > z^2 -- see if you can justify this -- but that's backwards logic.)

--

in any case, let's take the two statements together.
it's obvious that you can get a YES answer to the question; all you have to do is take ridiculously big numbers for x and y, and a small number for z. for instance, x = y = 100, z = 0, satisfy both statements, and clearly give a YES answer.

so, you're trying for a NO answer.

try to make Z as big as possible while still satisfying the criteria (i.e., less than x^2 + y^2).
let's let x = y = 3
then to satisfy both statements, we need z^2 less than 18, and z less than 6.
we'll take z = 4, which is pushing the limit of the first one.
in this case, then, x^4 + y^4 = 81 + 81 = 162, but z^4 = 256, giving a NO answer.

insufficient
answer = e

Hi Ron,

Could you explain a bit further on this :

that if you KNOW x^4 + y^4 > z^4, then you know conclusively that x^2 + y^2 > z^2 -- see if you can justify this -- but that's backwards logic.)

RON RON RON RON RON for president!

if you know that x^4 + y^4 > z^4, then you can add the term 2(x^2)(y^2) to the left-hand side, which is already greater. because 2(x^2)(y^2) is nonnegative, the left-hand side will still be greater:
x^4 + 2(x^2)(y^2) + y^4 > z^4
because every term in this inequality carries an even exponent, we know that all the terms are nonnegative. therefore, we can square root both sides to yield
x^2 + y^2 > z^2
done.