Ukrainian Mathematical Olympiad

Let $u=a+b$ and $v=\frac{1}{a}+\frac{1}{b}$. Then $u+v=5$. As is known, $uv\ge 4$. Therefore, $5=u+v\ge u+\frac{4}{u}$. Hence, $u^2-5u+4\le 0$. Thus, $1\le \sqrt{u}\le 2$, that is, $(\sqrt{u}-1)(\sqrt{u}-2)\le 0$, and $3\sqrt{u}\ge u+2$.