= Solution
For the equality-constrained <minimization problem>, $y=8/x$ and $x\ne0$. Put $t=x^2>0$. The restricted objective is
$$
f(x,8/x)=t^2+\frac{128}{t}-28=:F(t),\qquad F'(t)=2t-\frac{128}{t^2}.
$$
The only zero has $t^3=64$, so $t=4$. Since $F''(t)=2+256/t^3>0$, the <second-derivative criterion for convexity> proves a unique minimizing $t$. Consequently \b[the equality constraint has minimizers $(2,4)$ and $(-2,-4)$, with minimum $20$].
For the inequality-constrained <minimization problem>, the identity $f=(x^2-1)^2+2(y-x)^2+3$ supplies the universal lower bound $3$. Both equality points for that bound, $(1,1)$ and $(-1,-1)$, satisfy $xy=1\le8$. Hence \b[the inequality constraint has exactly these two minimizers, with minimum $3$].
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