examday

/TMUA

Let $a$ and $b$ be distinct positive real numbers satisfying

A student gives the following proof that $a+b>2$.

1. Multiplying the given equation by $ab$ gives $a^2b+b=ab^2+a$.
2. Rearranging and factorising gives $(a-b)(ab-1)=0$.
3. Since $a\ne b$, it follows that $ab=1$.
4. Therefore $(a+b)^2=(a-b)^2+4ab>4$.
5. Since $a+b>0$, this gives $a+b>2$.

Which of the following is correct?