If $\mathbf{b}=0$, the objective is zero at every feasible point, so every point of the ellipsoid is a minimizer. Suppose now $\mathbf{b}\ne 0$. A minimum cannot be interior: a sufficiently small step in direction $-\mathbf{b}$ would remain feasible and decrease the objective. The boundary is active.

Forstationarity gives $\mathbf{b}+2\lambda A\mathbf{x}=0$. A positive-definite symmetric matrix is invertible, and the minimizing sign has $\lambda >0$. Enforcing the boundary gives

For global optimality, use the inner product $⟨u,v{⟩}_A=u^{\mathsf{T}}Av$. The Cauchy-Schwarz inequality givesThe proposed point attains the negative of this bound. Thus the minimum is $-\sqrt{{\mathbf{b}}^{\mathsf{T}}{A}^{-1}\mathbf{b}}$, and equality in the Cauchy-Schwarz inequality, together with the sign and boundary conditions, makes this minimizer unique when $\mathbf{b}\ne 0$.

Use the LagrangianThe Lagrange multiplier equations areFeasibility makes $x\ne 0$, so $\lambda =2x^2>0$. The second equation becomes $2y(8y^2+\lambda )=0$, forcing $y=0$ and then $x=\pm 1$. The constraint gradient $(2x,-2y)$ is nonzero on the feasible set, so the multiplier calculation has no missing singular feasible point.

To verify the global minimum rather than just stationarity, eliminate $x^2=1+y^2$:Equality holds exactly at $(x,y)=(1,0)$ or $(-1,0)$, with minimum value $1$.

For the vector equation, the Jacobian matrix isWrite the updated coordinates as $(X,Y)$. The second equation of the multivariable Newton iteration gives $X-Y=0$. Substituting $X=Y$ into its first linearized equation givesHence $h(x,y)=6x^5-20y^3$, and, whenever $h\ne 0$,

Starting from $(0,1)$ gives $(x_1,y_1)=(21/20,21/20)$. Every subsequent iterate remains on the diagonal, where this is exactly the scalar Newton root-finding iteration from (b). The value $f(21/20)>0$, so $21/20<r$.

For $1\le x<r$, concavity again puts the tangent's zero at or above $r$. Alsoso $r\le N(21/20)\le N(1)=8/7<6/5$. After one scalar step the sequence therefore enters the invariant interval proved in (b). It converges to $(r,r)$ at least quadratically. If the vector error is measured in the Euclidean norm, then $\Vert (x_k,y_k)-(r,r)\Vert =\sqrt{2}|x_k-r|$, so the scalar quadratic estimate also gives a quadratic vector estimate. The initial off-diagonal step causes no difficulty because its Jacobian matrix is invertible and lands exactly on the diagonal.