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On an open convex set, a twice-differentiable real-valued function is convex if and only if its Hessian matrix is a positive semidefinite matrix at every point. In particular continuous second partial derivatives suffice for the differentiability hypothesis. Restrict $g$ to any line segment: $\varphi (t)=g(x+tv)$ has ${\varphi }^{\prime \prime }(t)=v^{\mathsf{T}}H(g)(x+tv)v$. The second-derivative criterion for convexity proves both implications, because convexity is equivalent to convexity of all such restrictions.

Expand before taking the second derivative:The second-derivative criterion for convexity proves that $f$ is strictly convex, hence convex.

The second-derivative criterion for convexity isThe non-strict inequality is necessary and sufficient; the stronger condition $f^{\prime \prime }>0$ is sufficient for strict convexity.

The Hessian criterion for convexity, in the usual twice-differentiable multivariate setting, isThus the Hessian matrix must be a positive semidefinite matrix everywhere. The criterion follows by applying the second-derivative criterion for convexity to each line restriction $t\mapsto g(x+tv)$, whose second derivative is $v^{\mathsf{T}}H(g)(x+tv)v$. Continuous second partials are a sufficient regularity hypothesis for this line-restriction argument.

For the equality-constrained minimization problem, $y=8/x$ and $x\ne 0$. Put $t=x^2>0$. The restricted objective isThe only zero has $t^3=64$, so $t=4$. Since $F^{\prime \prime }(t)=2+256/t^3>0$, the second-derivative criterion for convexity proves a unique minimizing $t$. Consequently 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 $xy=1\le 8$. Hence the inequality constraint has exactly these two minimizers, with minimum $3$.