Prove that $5x^2−2xy−8x+ 2y^2−2y+ 5 \ge 0$ for all $x, y\in\mathbb R$. When does equality occur?

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.

• @HaiDangel29 One of the possible solutions involving discriminant doesn't mean the question needs to be tagged (discriminant)... – YuiTo Cheng May 25 at 9:18

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.

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

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

• very elaborated, thanks! – Andreas May 24 at 6:13

$$5x^2-2xy-8x+2y^2-2y+5=x^2-2xy+y^2+4x^2-8x+4+y^2-2y+1=(x-y)^2+(2x-2)^2+(y-1)^2$$

• So equality happens iff $x=y=1$. – RMWGNE96 May 23 at 19:45

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?

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 $$\because\,{\rm discriminant}[\,5\,x^{\,2}- 2\,xy- 8\,x+ 2\,y^{\,2}- 2\,y+ 5,\,x\,]= -\,36(\,y- 1\,)^{\,2}\leqq 0$$