2
$\begingroup$

Show that for $a,b \in \mathbb{R}$ such that $2b>1$ and $a>\sqrt{b}$ the following equality is true:

$$a^4-a^2>b^2-b.$$

I started as follows

$$a^2(a-1)(a+1)-b(b-1)>0.$$

I do not know what to do with this. I would be grateful for any help.

$\endgroup$

1 Answer 1

1
$\begingroup$

That's the same as $a^4-b^2 > a^2-b$ or $(a^2-b)(a^2+b) > a^2-b$ which makes it obvious.

Just in case:

$a^2+b > 2b > 1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .