[ThE-dArK-lOrD]

Solution for #5 (Cr. Sirius)
We denote $P(x,y)$ for $f(x)^2 \geq f(x+y)(f(x)+y)$ for all $x,y \in \mathbb{R}^+$
Since $f(x+y)(f(x)+y) >f(x)f(x+y)$
From $P(x,y)$ we get $f(x) > f(x+y)$ for all $x,y \in \mathbb{R}^+$, so $f$ is strictly decreasing
Then $P(x,f(x))$ give us new assertion $Q(x); \frac{f(x)}{2} \geq f(x+f(x))$ for all $x\in \mathbb{R}^+$
Since $Q(x+f(x)),Q(x)$ and $f$ is strictly decreasing give us $\frac{f(x)}{4} \geq \frac{f(x+f(x))}{2} \geq f(x+f(x)+f(x+f(x))) \geq f(x+f(x)+\frac{f(x)}{2})$ for all $x\in \mathbb{R}^+$
So we get assertion $R_1(x);\frac{f(x)}{4} \geq f(x+\frac{3}{2}f(x))$
From $R_1(x+f(x)),Q(x)$ and $f$ is strictly decreasing give us $\frac{f(x)}{8} \geq \frac{f(x+f(x))}{4} \geq f(x+f(x)+\frac{3}{2}f(x+f(x))) \geq f(x+f(x)+\frac{3}{2} \cdot \frac{f(x)}{2})=f(x+\frac{7}{4}f(x))$ for all $x\in \mathbb{R}^+$
So we get assertion $R_2(x); \frac{f(x)}{8} \geq f(x+\frac{7}{4}f(x))$
Then we can easily proved in similar way by induction on $n \in \mathbb{Z}^+$ that $R_n(x); \frac{f(x)}{2^{n+1}} \geq f(x+(2-\frac{1}{2^n})f(x))$ for all $x\in \mathbb{R}^+$
So we get that for all $n\in \mathbb{Z}^+$, we have $\frac{f(x)}{2^{n+1}} \geq f(x+(2-\frac{1}{2^n})f(x)) >f(x+2f(x))$ for all $x\in \mathbb{R}^+$
But when $n \rightarrow \infty$ we have $\frac{f(x)}{2^{n+1}} \rightarrow 0$
Contradict the existence of $f(x+2f(x)) \in \mathbb{R}^+$

P.S. Dear Amankris
I believe that you have beautiful solution by A.M.-G.M., can you show it to me, plz ?