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$$A = x^4 ( 1 + y^4 ) + y^4 ( 1 + z^4 ) + z^4 ( 1 + x^4 )$$ $$B = x^2 y^2 z^2$$

If $x$, $y$ and $z$ are real, which of the following can be the value of $A/B$?


i) 0
ii) 2
iii) 4
iv) 8

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Welcome to MathSE. I see that you are relatively new here. So I wanted to let you know a few things about MathSE. We like to know where the problem is from what you've tried on a problem; this prevents people from wasting their time telling you thinks you already know, and helps make sure the answers are at an appropriate level. If this is homework, please consider adding the [homework] tag; people will still help, so don't worry. – Arturo Magidin Feb 27 '12 at 6:20
@Arturo i will do that.sorry. – vikiiii Feb 27 '12 at 6:47
up vote 2 down vote accepted


Expand $A$ to be a sum of $6$ terms and use $\text{AM} \ge \text{GM}$.

For the sake of completeness, we add more detail.

Since we talk of the ratio $\frac{A}{B}$, we can assume that $B \neq 0$, and thus none of $x,y,z$ are non zero.

Now $A = x^4 + x^4y^4 + y^4 +y^4z^4 + z^4 + z^4x^4 \ge 6 x^2y^2z^2 = 6B$, using $\text{AM} \ge \text{GM}$.

Thus the ratio $\frac{A}{B}$ is atleast $6$. So if there is an answer, it must be iv) 8. Note, we can eliminate $0$, as we need $B \neq 0$.

Extra Credit:

We now show that for any $r \ge 6$, there are $x,y,z$, such that we have $\frac{A}{B} = r$.

We arbitrarily choose $y = z = 1$.

This gives us the equation

$3x^4 + 3 = r x^2$.

This is a quadratic in $x^2$ and has the positive root $\frac{r + \sqrt{r^2 - 36}}{6}$.

Thus we have that $\frac{A}{B}$ can only take values $\ge 6$, and for each such value, there are $x,y,z$ for which we attain that ratio.

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Thanks Aryabhata.Got it .It will be 6 . :) – vikiiii Feb 27 '12 at 6:47
@vikiiii: It will be atleast $6$. – Aryabhata Feb 27 '12 at 6:48

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