Solution of Non Linear System of Inequalities

Question

The given question is :
Consider the sets defined by the real solutions of the inequalities
$A=\left \{(x,y):x^2+y^4\leqslant 1 \right \}\; B=\left \{(x,y):x^4+y^6\leqslant 1 \right \} $
A. $B\subseteq A$
B. $A\subseteq B$
C. Each of the sets $A–B$,$B–A$ and $A\cup B$ is non-empty
D. None of the above

Now I found that the answer will be Option (B) by using an online graph plotter but my question is how to solve this without using any such plotter.I was thinking about using the methods to solve non-linear system of equations but could not figure out how to do it for inequalities.Please help and if any other method is there to solve this do mention.

Answer 1

Note that if $|x|>1$ or $|y|>1$, then $(x,y)$ does not belong to $A$ or $B$. Suppose $(x,y)\in A$, so that $x^2+y^4\le 1$. Then we have that $$ x^4 + y^6 = x^2\times x^2 + y^2\times y^4 \le 1^2\times x^2 + 1^2\times y^4 = x^2 + y^4\le 1. $$ It follows that $(x,y)\in B$ and thus $A\subseteq B$.

Answer 2

All points $(x,y)$ in $A$ or $B$ must satisfy $|x| \leq 1$ and $|y| \leq 1$. Under these conditions, you have that $x^4 \leq x^2$ and $y^6 \leq y^4$. So, $$ x^2 + y^4 \leq 1 \Rightarrow x^4 + y^6 \leq 1, $$ meaning that every point in $A$ will also be in $B$, thus $A \subseteq B$.