3
$\begingroup$

Let $a,b\geq 1/2$. Prove that $$a^2b^2+2(a+b)\geq 4ab+1.$$

We know that $(ab-1)^2\geq 0$ implies $a^2b^2+1\geq 2ab$, so the inequality reduces to $2(a+b)\geq 2ab+2$, or $a+b\geq ab+1$. But this is equivalent to $(a-1)(b-1)\leq 0$, which is not true. How can we fix it?

$\endgroup$
  • $\begingroup$ The slightly (but only slightly) flippant answer is: Don't do that, then. To be less flippant, what you have discovered is that the inequality you used is to rough an estimate to work. So you'll need to abandon that approach and try something different. $\endgroup$ – Harald Hanche-Olsen Dec 17 '14 at 15:51
3
$\begingroup$

Hint: $(2a-1)(2b-1)=4ab-2(a+b)+1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.