Question 84
Let $x$, $y$ and $z$ be positive real numbers such that $x^4 + y^4 + z^4 = 3$. Prove that
$$
\frac{9}{x^2 + y^4 + z^6} + \frac{9}{x^4 + y^6 + z^2} + \frac{9}{x^6 + y^2 + z^4} \le x^6 + y^6 + z^6 + 6.
$$
When does equality hold?
$$
\frac{9}{x^2 + y^4 + z^6} + \frac{9}{x^4 + y^6 + z^2} + \frac{9}{x^6 + y^2 + z^4} \le x^6 + y^6 + z^6 + 6.
$$
When does equality hold?