Does there exist the function $f^2(x)\ge f(x+y)\left(f(x)+y \right)$

Does there exist the function $f:\mathbb R^+\rightarrow \mathbb R^+$, such that $$f^2(x)\ge f(x+y)\left(f(x)+y \right) \forall x,y \in \mathbb R^+$$

My work so far:

Assume that a function exists. Then $$f(x+y)\le \frac{f^2(x)}{f(x)+y}<f(x).$$ Then this function is strictly decreasing. (And this function is injective).

• How did you get the second inequality? – sranthrop May 5 '16 at 12:50
• @sranthrop: I edited. – Roman83 May 5 '16 at 12:54

Taking logarithms gives $$2\log f(x)\geq\log f(x+y)+\log(1+y/f(x))+\log f(x)$$ After rearranging, $$\frac{\log f(x)-\log f(x+y)}{y}\geq\frac{\log(1+y/f(x))}{y},$$ so after taking limit as $y\rightarrow0$ we get $$\frac{-f'(x)}{f(x)}=(-\log f)'(x)\geq\frac{1}{f(x)}.$$ and so $f'(x)\leq-1$, which contradicts $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$. (As $f$ is decreasing, it is almost everywhere differentiable)
• $f$ may not derivative; – partofsha May 5 '16 at 13:27
• You both could combine your answers and simplify mge's idea a little: There is no need to take logarithms if you use partofsha's first inequality. Then you immediately get $f'(x)\leq -1$ a.e. – sranthrop May 5 '16 at 13:31
• Why $f'(x)\leq-1$ contradicts $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$? Let $f(x)=\frac 1 x$ – user261263 May 5 '16 at 15:28
Suppose that there is a function $f$.then we have since $$f^2(x)\ge f(x+y)(f(x)+y)$$ we have $$f(x)-f(x+y)\ge\dfrac{f(x)y}{f(x)+y}>0$$ which shows that $f$is a strictly decreasing function.Give an $x\in R^{+}$, we choose an $n\in N^{+}$ such $nf(x+1)\ge 1$, then $$f\left(x+\dfrac{k}{n}\right)-f\left(x+\dfrac{k+1}{n}\right)\ge\dfrac{f(x+k/n)\cdot\frac{1}{n}}{f(x+\frac{k}{n})+\frac{1}{n}}>\dfrac{1}{2n}$$ summing up these inequalities for $k=0,1,\cdots,n-1$,we have $$f(x)-f(x+1)>\dfrac{1}{2}$$ take an $m$ such $m\ge 2f(x)$,then $$f(x)-f(x+m)>\dfrac{m}{2}\ge f(x)$$a contradiction