Let $x$ and $y$ be real numbers. Consider $t=x^2+10y^2-6xy-4y+13$. So what is the $t$ as smallest number? The solution is $9$.
My trying:
$$t=x^2+10y^2-6xy-4y+13$$
$$=x^2+9+10y^2-6xy-4y+4$$ $$=(x+3)^2-3x+2y(5y-2)-6xy+4$$.
So let $x=-3$ and $y=0$. From this I got $t=13$. My answer is false. Can you help?
On
You need to get everything into squares or constants, since otherwise you can't minimise all the individual terms. The fact that you have an $xy$ term and a $y$ term, but no $x$ term, suggests you want something of the form $$(ax+by)^2+(cy+d)^2+e.$$ Can you find values of $a,b,c,d,e$ to make this work?
On
Notice that $$t=x^2+10y^2-6xy-4y+13=x^2-6y \cdot x+10y^2-4y+13,$$which could be regarded as a quadratic function with respect to the variable $x$.
Hence, $$t \geq \dfrac{4\cdot 1 \cdot (10y^2-4y+13)-(-6y)^2}{4 \cdot 1}=y^2-4y+13=(y-2)^2+9\geq 9,$$ with the equality holding if and only if $x=-\dfrac{-6y}{2 \cdot 1}$ and $y=2$,namely, $x=6,y=2$.
Use Gauß' method to write a quadratic form as a linear combination of squares of linear forms: \begin{align} t&=x^2+10y^2-6xy-4y+13 =(x-3y)^2 -9y^2+10y^2-4y+13 \\ &=(x-3y)^2+y^2-4y+13 =(x-3y)^2+(y-2)^2-4+13\\ &=(x-3y)^2+(y-2)^2+9. \end{align}