This is a strengthened version of 2007 Mediterranean Mathematical Olympiad, Problem 4. Let $x=n+r$ where $n=\lfloor x\rfloor\in\mathbb{N}$ and $r=\{x\}\in(0,1)$. After some algebra, the inequality boils down proving \[ \frac{4r(n+r)}{n(n+2r)} + \frac{4n(n+r)}{r(2n+r)} = \frac{4(n+r)^2(n^2+nr+r^2)}{nr(2n+r)(n+2r)}>\frac{16}{3}. \]Next, using the fact \[ (2n+r)(n+2r)\le \frac{9}{4}(n+r)^2 \]valid from AM-GM, it suffices to prove \[ n^2+nr+r^2>3nr \Leftrightarrow (n-r)^2>0, \]which is clearly true. The constant $16/3$ is essentially best possible, which can be seen by taking $x=2-\epsilon$ for $\epsilon>0$ arbitrarily small.