# Inequality $a^2b^2+2(a+b)\geq 4ab+1$

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?

• 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. – Harald Hanche-Olsen Dec 17 '14 at 15:51

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