# Proof: $x (y - x + 1) > y$ if $y > x$

I'm working on a computation which depends on the idea that given two natural numbers $x$ and $y$ where $y > x$, the product $x(y - x)$ will always be greater than $y$.

Is there a proof of this ? My elementary math is a bit rusty.

The simple evaluation gives : $xy - x^2$

I can't seem to formalize this relation with respect to $y$. Could somebody give me a refresher on the proof strategies for such a problem ?

*EDIT: * Apologies. I got lost in writing it here. It's actually $x (y - x + 1)$. So basically given two numbers this would result in the value above plus a summation series that is solvable using $x(x+1)/2$. Does that make more sense ? (P.S: thank you for the quick response.)

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Let $y = 2$ and $x = 1$. Then $y > x$, but $x (y - x) = 1 \cdot (2-1) = 1$ and $y = 2$. Your statement is wrong. – k.stm Jan 20 '13 at 10:53

$$x(y-x+1)>y\iff xy-y>x(x-1)$$ $$\iff y(x-1)>x(x-1)\iff (y-x)(x-1)>0$$

which will be true if $(y>x$ and $x>1)$ or if $(y<x$ and $x<1)$

i.e., if $y>x>1$ or $y<x<1$

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Yippee... I just saved a variable cause of this. Thanks. – Saad Farooq Jan 20 '13 at 11:27
@SaadFarooq, my pleasure. – lab bhattacharjee Jan 20 '13 at 11:28

Try $y=x+1$, then the product is $x$, which is not greater than $y$. Hence you try to prove something wrong.

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Try $x=2$ and $y=3$. Then $x(y-x)=2<y$.

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A different way to look at the problem: $$x(y-x+1)=-x^2+(y+1)x$$ with fixed $y$ is a quadratic function in $x$. The coefficient by $x^2$ is negative, so the maximal value of it will be at $x=\frac{y+1}{2}$, which is smaller than $y$ if $y>1$, and the value is equal to $\frac{(y+1)^2}{4}$, which is clearly larger than $y$ for $y\geq 3$, for example.

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