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How to proof simply that $$ \forall x,y \in \mathbb{R} \qquad x^2+y^2+(x-1)(y-1)>0 $$

Hamza
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    Have you tried the problem? If you could, please show some of your own work. –  Nov 25 '14 at 01:50
  • Yes I can proof it by two ways (the first i use functions and the second i use a discussion about the values of x and y) but i search the most simple way. – Hamza Nov 25 '14 at 01:55
  • You can use mathematical induction to support your proof. –  Nov 25 '14 at 02:00
  • Please take a look at Rory Daulton's solution and comment thereafter. The lower bound is $2/3$. – mike Nov 25 '14 at 02:42

4 Answers4

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$$x^2+y^2+(x-1)(y-1) = x^2 + y^2 + xy - x -y+1 = \dfrac{(x+y)^2}2 + \dfrac{(x-1)^2}2 + \dfrac{(y-1)^2}2$$

Adhvaitha
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The left side of your inequality is a quadratic form with non-zero $xy$ term. Rotate the coordinates by 45 degrees and you will get an easily recognized form for an ellipse. So substitute $$x=x'+y'$$ $$y=x'-y'$$ to get an easier form that will be easy to prove to be positive. In fact, you will get a better lower bound for your expression than zero.

Rory Daulton
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$x^2+y^2+(x-1)(y-1)>0\iff x^2+y^2+xy-x-y+1>0\iff2x^2+2y^2+2xy-2x-2y+2>0\iff(x-1)^2+(y-1)^2+(x+y)^2>0$ which is obviously true.

Landon Carter
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You onle need to show that:

$$x^2+(y-1)x+y^2-y+1 > 0.$$

Try to show that $\Delta < 0$

Paul
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