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I tried grouping the $x$'s and the $y$'s but that didn't get me anywhere. I know that $5x^2, 2y^2$, and $5$ are always positive. I am not sure what to try next.

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  • $\begingroup$ @HaiDangel29 One of the possible solutions involving discriminant doesn't mean the question needs to be tagged (discriminant)... $\endgroup$ – YuiTo Cheng May 25 at 9:18
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By inspection, equality occurs for $x=y=1$.

So the obvious choice is to group around these values. This also gives a somewhat more detailed explanation how you arrive at these magic squares:

$$ 5x^2−2xy−8x+ 2y^2−2y+ 5 = \\ 5((x-1) +1)^2−2((x-1)+1)((y-1) + 1)−8(x-1)+ 2((y-1)+1)^2−2(y-1) -5\\ = 5(x-1)^2 - 2(x-1)(y-1) +2(y-1)^2 =\\ = 4(x-1)^2 + ((x-1)-(y-1))^2 +(y-1)^2 = \\ = 4(x-1)^2 + (x-y)^2 +(y-1)^2 \\ \geq 0 $$ which is clear.

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take the expression $$5x^{2}-2xy-8x+2y^{2}-2y+5$$, we want to write this as a sum of squares somehow. grouping the terms as: $$(x^{2}-2xy+y^{2}) + (y^{2}-2y+1) + (4x^{2}-8x+4) $$ factoring the three expressions gives $$(x-y)^{2} + (y-1)^{2} + 4(x-1)^{2}$$ which is the sum of squares, each of which is greater than $0$. for equality to hold, notice that both $(y-1)^{2}$ and $4(x-1)^{2}$ have roots of 1 which means they will be zero at that value, and also notice that $(x-y)^{2}$ will be zero whenever $x=y$, this means that we should take $x=y=1$ for equality to hold

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Bit late to the party. A sort of minimalist expression is $$ \frac{1}{5}(5x-y-4)^2 + \frac{9}{5} (y-1)^2 $$ so we get zero only when $y=1$ and $5x-1-4 =0,$ so also $x=1.$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 5 } & 1 & 0 \\ - \frac{ 4 }{ 5 } & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 5 } & - \frac{ 4 }{ 5 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & - 1 & - 4 \\ - 1 & 2 & - 1 \\ - 4 & - 1 & 5 \\ \end{array} \right) $$

for the curious, there is an algorithm for symmetric matrices:

Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 5 & - 1 & - 4 \\ - 1 & 2 & - 1 \\ - 4 & - 1 & 5 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 5 & - 1 & - 4 \\ - 1 & 2 & - 1 \\ - 4 & - 1 & 5 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 5 & 0 & - 4 \\ 0 & \frac{ 9 }{ 5 } & - \frac{ 9 }{ 5 } \\ - 4 & - \frac{ 9 }{ 5 } & 5 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & \frac{ 4 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 5 } & \frac{ 4 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 5 } & - \frac{ 4 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & - \frac{ 9 }{ 5 } \\ 0 & - \frac{ 9 }{ 5 } & \frac{ 9 }{ 5 } \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 5 } & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 5 } & - \frac{ 4 }{ 5 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 5 } & 1 & 0 \\ 1 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & - 1 & - 4 \\ - 1 & 2 & - 1 \\ - 4 & - 1 & 5 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 5 } & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 5 } & 1 & 0 \\ - \frac{ 4 }{ 5 } & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & \frac{ 9 }{ 5 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 5 } & - \frac{ 4 }{ 5 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & - 1 & - 4 \\ - 1 & 2 & - 1 \\ - 4 & - 1 & 5 \\ \end{array} \right) $$

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  • $\begingroup$ very elaborated, thanks! $\endgroup$ – Andreas May 24 at 6:13
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$5x^2-2xy-8x+2y^2-2y+5=x^2-2xy+y^2+4x^2-8x+4+y^2-2y+1=(x-y)^2+(2x-2)^2+(y-1)^2$

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  • $\begingroup$ So equality happens iff $x=y=1$. $\endgroup$ – RMWGNE96 May 23 at 19:45
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The LHS may be written as $$4x^2−8x+4\; +\; x^2−2xy+y^2\; +\; y^2−2y+ 1$$ Do you see why this yields the "$\,\ge 0\,$" ?
And also the equality case?

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We have $$5\,x^{\,2}- 2\,xy- 8\,x+ 2\,y^{\,2}- 2\,y+ 5= (\,1+ x- 2\,y\,)(\,5\,x+ 8\,y- 13\,)+ 18(\,y- 1\,)^{\,2}$$ $$9(\,5\,x^{\,2}- 2\,xy- 8\,x+ 2\,y^{\,2}- 2\,y+ 5\,)= -\,(\,1+ x- 2\,y\,)(\,5\,x+ 8\,y- 13\,)+ 2(\,5\,x- y- 4\,)^{\,2}$$ $$\therefore\,5\,x^{\,2}- 2\,xy- 8\,x+ 2\,y^{\,2}- 2\,y+ 5\geqq 0$$ Furthermore $$\because\,{\rm discriminant}[\,5\,x^{\,2}- 2\,xy- 8\,x+ 2\,y^{\,2}- 2\,y+ 5,\,x\,]= -\,36(\,y- 1\,)^{\,2}\leqq 0$$

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