<s>[INST] Find the minimum value of $9x^2+2y^2-8xy-6x+11$ For any $x, y$ in real, find the minimum value of $9x^2+2y^2-8xy-6x+11$

 [/INST] Let $$9x^2+2y^2-8xy-6x+11=k$$
Thus, the the following quadratic equation of $y$
$$2y^2-8xy+9x^2-16x+11-k=0$$ has real solutions, which says
$$16x^2-2(9x^2-6x+11-k)\geq0$$ or
$$(x-3)^2\leq k-2,$$ which gives $$k\geq2.$$
The equality occurs for $x=3$ and $y=6,$ which says that $2$ a minimal value.
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