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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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up vote 3 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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