1. ## Proving

How do you prove that x^2+y^2-2xy+x-y+1 is greater than 0?

2. ## Re: Proving

You can't.

\displaystyle \displaystyle \begin{align*} x^2 - 2\,x\,y + y^2 + x - y + 1 &= \left( x - y \right)^2 + x - y + 1 \end{align*}

All that you can say for sure is that \displaystyle \displaystyle \begin{align*} \left( x - y \right)^2 \end{align*} is nonnegative. The entire quantity will only be positive if \displaystyle \displaystyle \begin{align*} x - y + 1 > 0 \end{align*}, which you are not guaranteed.

3. ## Re: Proving

Originally Posted by Prove It
You can't.

\displaystyle \displaystyle \begin{align*} x^2 - 2\,x\,y + y^2 + x - y + 1 &= \left( x - y \right)^2 + x - y + 1 \end{align*}

All that you can say for sure is that \displaystyle \displaystyle \begin{align*} \left( x - y \right)^2 \end{align*} is nonnegative. The entire quantity will only be positive if \displaystyle \displaystyle \begin{align*} x - y + 1 > 0 \end{align*}, which you are not guaranteed.
It does not matter if $\displaystyle x-y+1$ is negative; $\displaystyle (x-y)^2$ may be big enough to make the whole expression positive.

Indeed, $\displaystyle x^2+y^2-2xy+x-y+1=(x-y)^2+(x-y)+1=\left(x-y+\frac12\right)^2+\frac34$ is clearly always positive.

4. ## Re: Proving

Excellent proof ...My congratulations to Nehushtan

Minoas