IRL_ABooklet

If $x,y\ge 0$, we can use a 4-term AM-GM to obtain $$x^4+y^4+18=x^4+y^4+9+9\ge 4(x^4\cdot y^4\cdot 9\cdot 9{)}^{1/4}=12xy.$$ We get equality if and only if all four terms are equal, i.e. if and only if $x=y=\sqrt{3}$. If one of the two numbers $x,y$ is non-positive and the other non-negative, the RHS of the target inequality is non-positive whereas the LHS is at least 18, so the inequality still holds, but with no possibility of equality. Finally, if $x,y<0$, we replace $(x,y)$ by $(-x,-y)$ which leaves both LHS and RHS unchanged, so the inequality holds by what we have shown already, and equality iff $x=y=-\sqrt{3}$. In summary, we have proved the inequality and shown that equality occurs if and only if $x$ and $y$ are both equal to $\sqrt{3}$ or both equal to $-\sqrt{3}$.

$$\begin{array}{rl}6(x-y{)}^2& =6x^2+6y^2-12xy,\\ (x^2-3{)}^2& =x^4-6x^2+9\text{and}\\ (y^2-3{)}^2& =y^4-6y^2+9,\end{array}$$ we have $$x^4+y^4+18-12xy=6(x-y{)}^2+(x^2-3{)}^2+(y^2-3{)}^2\ge 0$$ with equality iff $x=y=\pm \sqrt{3}$.