An indefinite quadratic has a unique boundary trust-region minimizer

Set $p_{\ast }=(1,1)\mathrm{/}\sqrt{2}$. Then $\mathrm{\parallel }{p}_{\ast }{\mathrm{\parallel }}_2=1$, $(B+2I)p_{\ast }=(1,3)\mathrm{/}\sqrt{2}=-g$, and $B+2I=\mathrm{diag}(1,3)$ is positive definite. With $\lambda =2$, complementarity holds at the boundary, so [F2] gives a global minimizer. For any other feasible $z\mathrm{\ne }{p}_{\ast }$, expansion gives $m(z)-m(p_{\ast })=\frac{1}{2}(z-p_{\ast }{)}^T(B+2I)(z-p_{\ast })+(1-\mathrm{\parallel }z{\mathrm{\parallel }}_2^2)>0,$ because the first term is strictly positive and the second nonnegative; hence the minimizer is unique.

The global minimizer has $g^T{p}_{\ast }=-2$ and $p_{\ast }^{T}B{p}_{\ast }=0$, so its predicted reduction is $2$. For the Cauchy step, $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=5$ and $g^{T}Bg=4$, hence [F3] gives $p_C=(1,3)\mathrm{/}\sqrt{10}$ and $\mathrm{\parallel }{p}_C{\mathrm{\parallel }}_2=1$. Its model value is $m(p_C)=f(x)-\sqrt{5}+2\mathrm{/}5$, so its predicted reduction is $\sqrt{5}-2\mathrm{/}5<2$; the strict inequality follows from $\sqrt{5}<12\mathrm{/}5$.