Prove that if $x > 3$ and $y < 2$, then $x^2 − 2y > 5$

Question

Suppose $x > 3$ and $y < 2$. Then $x^2 − 2y > 5$.

My attempt:

1. $x>3 \implies x^2 > 9$

2. $ y < 2 \implies 5 + 2y < 9$

It follows that $5+2y < 9 < x^2$. From this we can see that $5+2y < x^2 \implies x^2 -2y > 5$

Therefore, if $x>3$ and $y<2$, then $x^2 − 2y > 5$.

Is it accurate?

Answer 1

Yes, your proof is correct. Here is an alternative:

$x^2> 9$ and since $-y>-2$ we have $-2y>-4$, so $$x^2-2y>9-4 =5$$

Answer 2

Proof is fine:

Option:

1)$x >3$, then

$x^2= xx >3x >9$;

2) $y <2$, then

$x^2 - 2y >9-2y >9-2 \cdot 2=5$.