examday

/TMUA

Let $a$ and $b$ be positive real numbers satisfying $a+b+ab=3$, and write $s=a+b$.

A student gives the following proof that $ab\leq1$.

1. Since $(a-b)^2\geq0$, we have $s^2\geq4ab$.
2. Substituting $ab=3-s$ gives $s^2+4s-12\geq0$.
3. Hence $(s-2)(s+6)\geq0$, so $s\leq-6$ or $s\geq2$.
4. Since $s>0$, it follows that $s\geq2$.
5. Therefore $ab=3-s\leq1$.

Which of the following is correct?