Old gem. Note that as $f(x)+y>f(x)$, we trivially obtain that $f$ is decreasing. Next, we get \[ f(x+y)\le \frac{f(x)^2}{f(x)+y} = f(x)\left(1-\frac{y}{f(x)+y}\right)\iff f(x)-f(x+y)\ge \frac{f(x)y}{f(x)+y}. \]We now iterate this identity by taking $x=t+i/n$, $0\le i\le n-1$ and $y=1/n$: \[ f\left(t+\frac{i}{n}\right)-f\left(t+\frac{i+1}{n}\right)\ge \frac{nf(t+1)}{1+nf(t+1)}, \]using the fact $f$ is decreasing. Summing these up along $0\le i\le n-1$, we obtain \[ f(t)-f(t+1)\ge \frac{nf(t+1)}{1+nf(t+1)}, \]for all $n$. Keeping $t$ fixed and letting $n\to\infty$, we recover $f(t)-f(t+1)\ge 1$. So, $f(t)-f(t+N)\ge N$ for all $t>0$ and $N\in\mathbb{N}$. As $f(t)>f(t)-f(t+N)$, we thus obtain $f(t)>N$ for any $t,N$. This is an obvious contradiction by taking $N$ large enough.

If we suppose that $f(x+y) \geq f(x)$ for some $x$ and $y$, then $f(x)^2 \geq f(x+y)(f(x)+y) \geq f(x)(f(x) + y) = f(x)^2 + yf(x)$, i.e. $yf(x) \leq 0$, absurd. Hence $f$ is strictly decreasing. Now let $y=f(x)$ - then $f(x)^2 \geq f(x+f(x))(2f(x))$, i.e. $f(x+f(x)) \leq \frac{f(x)}{2}$. Oopsie, this is IZHO 2016 P4 for $k=2$, so we are done!