<s>[INST] 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 $y ≠ 0$.
 [/INST] 
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 $5^2 \ge 5^2-(-9)^2$.
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