Proving inequality given some conditions I want to show that $-x^2y^2-y^4+4y^2 \geq 0$, if $x^4+2y^2 \leq 4$ and $-x^2y^2-y^4+4y^2 \leq 0$, if $x^4+2y^2 \geq 32$. Is there a factoring trick that helps to prove this explicitly?

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*For the first question you need to prove
$$
-2x^2-2y^2+8 \geq 0
$$
with the condition $-2y^2 \geq x^4 -4$. This condition implies $x^2 < 2$ since otherwise it can never be fulfilled. Combining the two gives the sharper $
-2x^2+x^4+4 \geq 0
$ or $(x^2-1)^2 +3\ge 0$, done.


*For the second question you need to prove
$$
-2x^2-2y^2+8 \leq 0
$$
with the condition $-2y^2 \leq x^4 -32$. This condition needs to be applied only for $x^2 < \sqrt{32}$ since otherwise it is fulfilled anyway. So for this case, we have  $
-2x^2+x^4-24 \leq 0
$ or $(x^2-1)^2 \le 25$, or $x^2 \le 6$, but this is fulfilled with  $x^2 < \sqrt{32} < 6$.
Consider the other case  $x^2 \ge \sqrt{32}$, then we see directly that $
-2x^2-2y^2+8 \leq -2x^2 +8 \leq - 2 \sqrt{32} +8 < 0
$
