How to prove that if $x>0$, $y>0$, and $x>y$, then $x^2>y^2$? [duplicate]
Solution 1:
Hint: use the fact that $x^2 - y^2 = (x+y)(x-y)$.
How to prove that if $x>0$, $y>0$, and $x>y$, then $x^2>y^2$? [duplicate]
Solution 1:
Hint: use the fact that $x^2 - y^2 = (x+y)(x-y)$.