Is there a proof that $ x^2 > (x-y)(x+y)$? I noticed that when I did the math, $5^2 > 4 \cdot 6$, and $10^2$ is greater than $9 \cdot 11$, etc. I looked for a proof of this and couldn't find one. Assuming $x$ and $y$ are real numbers and $y ≠ 0$.
 A: Observe that by expanding out the brackets, for any real $x, y$, we have $$\begin{align}(x-y)(x+y)&=x^2-xy+xy-y^2 \\
&=x^2-y^2\\
&\le x^2\end{align}$$ since $y^2\ge 0$.
In particular, if $x, y$ are non-zero, then the inequality is strict since $y^2>0$.
A: We have $ (x-y)(x+y) = x^2 - y^2$, which is indeed smaller than $x^2$ when $y \neq 0$ since we then have $y^2 > 0$.
A: hint: compare $x^2$ sand $x^2-y^2$, where $y^2 \geq 0$
A: Counterexample:
Assuming that $x$ and $y$ are real numbers then then let $x=1$ and $y=0$ then $1^2>1^2-0^2=1$ a contradiction. So it cannot be proven just like that. You need to restrict $x$ and $y$.
A: 
For all $y > 0$, we have $y^2 > 0$ for all $y \not= 0$, 
  \begin{align}
x^2 &> x^2-y^2 =(x+y)(x-y)
\end{align}

I see that you also did some sample arithmetic with $x^2 \ge (x+y)(x-y)$. But try also testing $$x^2 > x^2-y^2$$ as well. E.g. $10^2 > 10^2-1 ^2$ or $5^2 \ge 5^2-(-9)^2$.
A: $$x^2>(x-y)(x+y)\\x^2> x^2-y^2\\x^2-(x^2-y^2)>0\\y^2=|y|^2>0$$
Which is clearly true for any $y\in \mathbb{R}- \{0\}$
