2016AC-3-3 - Rhea

Solution

Let $ t_1=x_1-2 $, $ t_2=x_2+1 $
so that $ g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0 $ would have some convex property
with $ f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1 $
$ D^2g(x)=\dfrac{1}{(t_1^2+t_2^2+3)^3}\begin{bmatrix} 6(t_1)^2-2(t_2)^3-6 & 8t_1t_2 \\ 8t_1t_2 & 6(t_2)^2-2(t_1)^3-6 \end{bmatrix} $

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