If $f(x_1) = f(x_2)$, comparing $P(x_1, y_0)$ and $P(x_2, y_0)$ gives us $\left(\frac{f(y_0)}{y_0}-1\right)(x_1-x_2) = 0 \implies x_1 = x_2 \implies f$ is injective.

Let $f(x) \le M$ for every $x \in \mathbb{R}^+$. From $P(x, y_0)$: \[ x = \frac{|f(f(x)+y_0)-f(x)-y_0|}{|\frac{f(y_0)}{y_0}-1|}. \]The right-hand side is bounded by $\frac{y_0+2M}{|\frac{f(y_0)}{y_0}-1|}$, while the left-hand side is not. Therefore, $f$ is unbounded.

Let $F$ be the set of fix points of $f$, and let $\alpha \in F$. From $P(x, \alpha)$, we obtain $f(x) + \alpha \in F$. From $P(\alpha, f(y))$, we obtain: \[ f(\alpha + f(y)) - \alpha = \left(\frac{f(f(y))}{f(y)}-1\right)\alpha + f^{m+1}(y) \implies f(\alpha + f(y)) = \frac{f(f(y))\alpha}{f(y)} + f^{m+1}(y). \]Therefore, $\frac{f(f(y))\alpha}{f(y)} + f^{m+1}(y) = f(\alpha + f(y)) = \alpha + f(y) \implies f(y)-f^{m+1}(y) = \alpha\left( \frac{f(f(y))}{f(y)} - 1 \right)$. Since we can change $\alpha$ to another fixed point (such as $\alpha + f(x)$ for any $x \in \mathbb{R}^+$), therefore the factors must be zero, i.e. $f(y) = f(f(y)) = f^{m+1}(y)$. Now, by subtracting $P(x, x)$ from $P(f(x), x)$, we obtain: $\left(\frac{f(x)}{x}-1\right)(f(x)-x) = 0 \implies f(x) = x$.

$P(x)=x^4+2x^3+x^2+x-1$, $P(0)<0$ and $P(\frac{1}{2})>0$ Suppose $\alpha$ is the max root in$(0,\frac{1}{2} )$($\approx 0.48402$ in fact) set $k=0.49,k\in (\alpha ,\frac{1}{2}),k^4+2k^3+k^2+k>1$ and set $N$ is arbitary large positive number Part1: $\forall x \in \mathbb{R^+}$, $f(x)\ge x$ Proofelse, $\exists f(t)<t$ Case1: $\exists \frac{f(t)}{t} <1-k$ $f(x)=f(f(x)+t)+(1-\frac{f(t)}{t})x-f^m(t)$ $\forall x>\frac{f^m(t)}{1-k-\frac{f(t)}{t}}$ $f(x)=f(f(x)+t)+kx+(1-k-\frac{f(t)}{t})x-f^m(t)$ $>f(f(x)+t)+kx$ We also obtain $f(x)>kx$ and $\forall x>\frac{f^m(t)}{k(1-k-\frac{f(t)}{t})} \Rightarrow f(x)+t>kx>\frac{f^m(t)}{1-k-\frac{f(t)}{t}}$ So $f(f(x)+t)>k(f(x)+t)>k^2x$ We have proved $f(x)>f(f(x)+t)+kx>(k^2+k)x$ Do the silimar thing once more : $\forall x>\frac{f^m(t)}{k^2(1-k-\frac{f(t)}{t})},f(x)+t>\frac{f^m(t)}{k(1-k-\frac{f(t)}{t})}$ $f(f(x)+t)>(k^2+k)(f(x)+t)>(k^2+k)f(x)>(k^4+2k^3+k^2)x$ $f(x)>f(f(x)+t)+kx>(k^4+2k^3+k^2+k)x>x$ $\forall x>\frac{f^m(t)}{k^3(1-k-\frac{f(t)}{t})},f(x)+t>\frac{f^m(t)}{k^2(1-k-\frac{f(t)}{t})}$ $\therefore f(f(x)+t)-f(x)>(k^4+2k^3+k^2+k-1)f(x)>(k^4+2k^3+k^2+k-1)x$ $\therefore (\frac{f(t)}{t}-1)x+f^m(t)>(k^4+2k^3+k^2+k-1)x$ $f^m(t)>(k^4+2k^3+k^2+k-\frac{f(t)}{t})x$ Let$x>\frac{f^m(t)}{k^4+2k^3+k^2+k-\frac{f(t)}{t}}$ leads to contradiction. Case2: $\forall x$ , $\frac{f(x)}{x}\ge 1-k$ and$1-k\le \frac{f(t)}{t}<1$ We will make it contradict $f(t)<t$ through the following approach , $\forall N>0$ a) $\exists x>N,f(x)<x$ b) $\forall d_2\in (0,1-\frac{f(t)}{t}),\exists x>N,\frac{f(x)}{x}<\frac{f(t)}{t}+d_2$ c) $\forall d_3\in (0,\frac{t}{f(t)}-1),\exists x>N,\frac{f(x)}{x}<2-\frac{t}{f(t)}+d_3$ d) $\forall n\in \mathbb{N^+},\forall d_4>0,\forall \frac{f(x)}{x}>\frac{n}{n+1}-d_4$ a) $\forall x>\frac{f^m(t)}{1-\frac{f(t)}{t}}$ , $f(f(x)+t)=f(x)-(1-\frac{f(t)}{t})x+f^m(t)<f(x)<f(x)+t$ and$f(x)+t>(1-k)x+t$ Additionally, let $x>\frac{N}{1-k}$ $\therefore f(x)+t>N ,f(f(x)+t)<f(x)+t$ i.e. $\exists x>N,f(x)<x$ b) Otherwise, $\forall d_2\in (0,1-\frac{f(t)}{t}),\exists N>0$ s.t. $\forall x>N,\frac{f(x)}{x}\ge \frac{f(t)}{t} +d_2$ . $\forall x>\frac{N}{1-k},f(x)+t>N$ $(\frac{f(t)}{t}-1)x+f^m(t)=f(f(x)+t)-f(x)$ $\ge (\frac{f(t)}{t}+d_2)(f(x)+t)-f(x)$ $>(\frac{f(t)}{t}+d_2-1)f(x)$ $\forall x>max\{\frac{N}{1-k},\frac{f^m(t)}{d_2}\}$ $(\frac{f(t)}{t}+d_2-1)f(x)<(\frac{f(t)}{t}-1)x+f^m(t)$ $<(\frac{f(t)}{t}-1+d_2)x$ $\Rightarrow f(x)>x$ for every large $x$ , contradiction. c) Otherwise, $\forall d_3\in (0,\frac{t}{f(t)}-1),\exists N>0,\forall x>N,\frac{f(x)}{x}\ge 2-\frac{t}{f(t)}+d_3$ $\forall x>\frac{N}{1-k},f(x)+t>N$ $(\frac{f(t)}{t}-1)x+f^m(t)=f(f(x)+t)-f(x)$ $\ge (2-\frac{t}{f(t)}+d_3)(f(x)+t)-f(x)$ $>(1-\frac{t}{f(t)}+d_3)f(x)$ Wish$(\frac{f(t)}{t}-1)x+f^m(x)<(1-\frac{t}{f(t)}+d_3)(\frac{f(t)}{t}+d_2)x$ Which lead to$\frac{f(x)}{x}>\frac{f(t)}{t}+d_2$ , and contradiction with b). And it$\Leftrightarrow (d_2+d_2d_3-\frac{d_2t}{f(t)}+\frac{d_3f(t)}{t})x>f^m(t)$ Thus , in b) pick $d_2<\frac{d_3(f(t))^2}{t^2},\forall x>max\{\frac{f^m(t)}{d_2+d_2d_3},\frac{N}{1-k}\}$ They respectively provides $(\frac{d_3f(t)}{t}-\frac{d_2t}{f(t)})x>0,(d_2+d_2d_3)x>f^m(t)$ the ineq desired is proven through adding them up. d) Prove it by induction: $n=1,\frac{f(t)}{t}>1-k>\frac{1}{2}>\frac{1}{2}-d_4$ when $n=k+1$ , if $\frac{f(x)}{x}\ge 1$ obviously true else, $\forall \frac{f(x)}{x}<1,\forall d_3,\exists x_0$ s.t. $\frac{f(x_0)}{x_0}<2-\frac{x}{f(x)}+d_3$ and we know from hypothesis $\frac{f(x_0)}{x_0}>\frac{k}{k+1}-d_4$ $\therefore 2-\frac{x}{f(x)}>\frac{k}{k+1}-d_3-d_4$ for $\forall d_3,d_4$ $\therefore 2-\frac{x}{f(x)}\ge \frac{k}{k+1}$ $\therefore \frac{f(x)}{x}\ge \frac{k+1}{k+2}$ $\therefore \forall n\in \mathbb{N^+}, f(t)<t,\frac{f(t)}{t}\ge \frac{n}{n+1}$ Let $n>\frac{\frac{f(t)}{t}}{1-\frac{f(t)}{t}}$ $\frac{f(t)}{t}<\frac{n}{n+1}$ , contradiction. $\therefore x\in \mathbb{R^+},f(x)\ge x$ If there exists $f(t)>t$ Part2: $\forall x\in \mathbb{R^+},f(x)>x$ Proofelse $\exists f(z)=z$ $\forall x,f(f(x)+z)=(\frac{f(z)}{z}-1)x+f(x)+f^m(z)=f(x)+z$ for $x>N,f(x)+z\ge x+z>N$ i.e. $\exists x>N,f(x)=x$ $f(f(x)+t)=f(x)+(\frac{f(t)}{t}-1)x+f^m(t)$ $\exists u>f^m(t)+t,f(u)=u$ Let $x=\frac{u-f^m(t)-t}{\frac{f(t)}{t}-1}$ $\therefore f(f(x)+t)+t=f(x)+u$ $f(u)=u\Rightarrow f(f(x)+u)=(\frac{f(u)}{u}-1)x+f(x)+f^m(u)=f(x)+u$ i.e. $ f(f(f(x)+t)+t)=f(f(x)+t)+t$ $t<f(t)<f^m(t)+(\frac{f(t)}{t}-1)(f(x)+t)$ $=f(f(f(x)+t)+t)-f(f(x)+t)=t$ Part3: $\forall x\in \mathbb{R^+},f(x)=x$ Proofa) denote $\frac{f(t)}{t}=r,\exists x>N,\frac{f(x)}{x}<\sqrt{r}+1-\frac{1}{\sqrt{r}}$ b) $\forall d>0,\exists x>N,\frac{f(x)}{x}<1+d$ c) $\forall d>0,\exists x>N,\frac{f^m(x)}{x}<1+d$ d) $\forall d,y>0,\forall x>\frac{f^m(y)}{d},\frac{f(f(x)+y)}{f(x)+y}<\frac{f(y)}{y}+d$ a) else, $\exists N,\forall x>N,\frac{f(x)}{x}\ge \sqrt{r}+1-\frac{1}{\sqrt{r}}$ $(r-1)x+f^m(t)=f(f(x)+t)-f(x)$ $\ge (\sqrt{r}+1-\frac{1}{\sqrt{r}})(f(x)+t)-f(x)>(\sqrt{r}-\frac{1}{\sqrt{r}})f(x)$ $>(\sqrt{r}-\frac{1}{\sqrt{r}})(\sqrt{r}+1-\frac{1}{\sqrt{r}})x$ $=(r+\frac{1}{r}+\sqrt{r}-\frac{1}{\sqrt{r}}-2)x$ $x<\frac{f^m(t)}{\frac{1}{r}+\sqrt{r}-\frac{1}{\sqrt{r}}-1}=\frac{f^m(t)}{(\sqrt{r}-1)(1-\frac{1}{r})}$ is a constant, contradiction. b) else $\forall d>0,\exists N>0,\forall x>N,\frac{f(x)}{x}\ge 1+d$ $\exists t_1>N,\frac{f(t_1)}{t_1}<\sqrt{r}+1-\frac{1}{\sqrt{r}}$ $(\frac{f(t_1)}{t_1}-1)<\frac{r-1}{\sqrt{r}}<(\frac{f(t)}{t}-1)\frac{1}{1+d}$ Repeating this will give us $\exists t_n,\frac{f(t_n)}{t_n}-1<\frac{r-1}{(\sqrt{1+d})^n}$ Pick $n$ s.t. $d(\sqrt{1+d})^n>r-1$ , contradiction. c) else $\forall d>0,\exists N>0,\forall x>N,\frac{f^m(x)}{x}\ge 1+d$ for a fixed $x_0$ , $\forall d'>0,\exists a>N+f(x_0),\frac{f(a)}{a}<1+d'$ $f(f(x_0)+a-f(x_0))-f(x_0)=(\frac{f(a-f(x_0))}{a-f(x_0)}-1)x_0+f^m(a-f(x_0))$ $>0+(1+d)(a-f(x_0))$ $(1+d)(a-f(x_0))<f(a)-f(x_0)<(1+d')a-f(x_0)$ $\Leftrightarrow (d-d')(a-f(x_0))<d'f(x_0)$ Let $d'<d,a>\frac{d'f(x_0)}{d-d'}$ , contradiction. d) else $\exists x>\frac{f^m(y)}{d}, \frac{f(f(x)+y)}{f(x)+y}\ge \frac{f(y)}{y}+d$ Since$\frac{f(f(x)+y)}{f(x)+y}=\frac{f(x)+(\frac{f(y)}{y}-1)x+f^m(y)}{f(x)+y}$ $f(x)+(\frac{f(y)}{y}-1)x+f^m(y)\ge (f(x)+y)(\frac{f(y)}{y}+d)$ $\therefore (\frac{f(y)}{y}-1)x+f^m(y)\ge (\frac{f(y)}{y}+d-1)f(x)>(\frac{f(y)}{y}+d-1)x$ $\Rightarrow f^m(y)>dx$ , contraditcion. In the end, we prove that if $f(t)>t,f(f(x)+t)-f(x)=(\frac{f(x_0)}{x_0}-1)x+f^m(t)$ cannot hold. Abstract?for $\frac{f(a)}{a}\rightarrow 1,\frac{f(f(x)+a)}{f(x)+a}=\frac{f(x)+(\frac{f(a)}{a}-1)x+f^m(a)}{f(a)+a}\rightarrow 1$ and in $f(f(x)+t)-f(x)=(\frac{f(t)}{t}-1)x+f^m(t)$ If $f(x)+t$ can be shown as $f(?)+a$ , we obtain $\frac{f(f(x)+t)}{f(x)+t}\rightarrow 1$ then $t\approx (\frac{f(t)}{t}-1)x+f^m(t)$ is linear about $x$ which could lead to contradition $\forall d_1,N_1>0,\exists a>N_1,\frac{f(a)}{a}<1+d_1$ let $x=f(p)+a$ undetermined , $y=t$ , then we have $f(f(f(p)+a)+t)=f(f(p)+a)+(\frac{f(t)}{t}-1)(f(p)+a)+f^m(t) \cdots (1)$ for the first term, let $x=p,y=a$ $f(f(p)+a)+t=f(p)+(\frac{f(a)}{a}-1)p+f^m(a)+t$ $\forall d_2>0,\exists b>f^m(a)+t,\frac{f^{\max\{m,1\}}(b)}{b}<1+d_2$ Let $p=\frac{b-f^m(a)-t}{\frac{f(a)}{a}-1}$ , $\therefore f(f(p)+a)+t=f(p)+b$ $\frac{p}{f^m(b)}>\frac{p}{b(1+d_2)}=\frac{b-f^m(a)-t}{b(1+d_2)(\frac{f(a)}{a}-1)}$ Let $d_2 +1<\frac{d_1}{\frac{f(a)}{a}-1} \Rightarrow \frac{p}{f^m(b)}>\frac{b-f^m(a)-t}{d_1b}$ $\forall d_3<\frac{1}{d_1}$ , let $b>\frac{f^m(a)+t}{1-d_1d_3} \Rightarrow \frac{p}{f^m(b)}>d_3$ and $b>(\frac{f(a)}{a}-1)d_3f^m(a)+f^m(a)+t$ $\Rightarrow \frac{p}{f^m(a)}=\frac{b-f^m(a)-t}{f^m(a)(\frac{f(a)}{a}-1)}>d_3$ $\forall x>d_3f^m(y),\frac{f(f(x)+y)}{f(x)+y}<\frac{f(y)}{y}+\frac{1}{d_3}$ We have $\frac{f(f(p)+a)}{f(p)+a}=\frac{f(f(p)+a)}{f(p)+a}<\frac{f(a)}{a}+\frac{1}{d_3}$ and $\frac{f(f(f(p)+a)+t)}{f(f(p)+a)+t}=\frac{f(f(p)+b)}{f(p)+b}<\frac{f(b)}{b}+\frac{1}{d_3}$ $\therefore (\frac{f(t)}{t}-1)(f(p)+a)=f(f(f(p)+a)+t)-f(f(p)+a)-f^m(t)$ $<(\frac{f(b)}{b}+\frac{1}{d_3})(f(f(p)+a)+t)-f(f(p)+a)-t$ $\le (\frac{f^{\max\{m,1\}}(b)}{b}+\frac{1}{d_3})(f(f(p)+a)+t)-f(f(p)+a)-t$ $<(1+d_2+\frac{1}{d_3})(f(f(p)+a)+t)-f(f(p)+a)-t$ $=(d_2+\frac{1}{d_3})f(f(p)+a)+(d_2+\frac{1}{d_3})t$ $<(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3})(f(p)+a)+(d_2+\frac{1}{d_3})t$ $\therefore (\frac{f(t)}{t}-1)(f(p)+a)<(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3})(f(p)+a)+(d_2+\frac{1}{d_3})t$ We require $(1+2d_1)d_1<\frac{f(t)}{t}-1,(d_2+d_1)(1+2d_1)<\frac{f(t)}{t}-1,$ $d_1<\frac12 ,2<d_3<\frac{1}{d_1} ,(d_2+\frac{1}{d_3})(1+d_1+\frac{1}{d_3})<\frac{f(t)}{t}-1$ $((\frac{f(t)}{t}-1)-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3}))(f(p)+a)<(d_2+\frac{1}{d_3})t<t$ Let $b>\frac{(\frac{f(a)}{a}-1)t}{\frac{f(t)}{t}-1-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3})}+f^m(a)+t$ $\Rightarrow p=\frac{b-f^m(a)-t}{\frac{f(a)}{a}-1}>\frac{t}{\frac{f(t)}{t}-1-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3})}$ $\therefore t<((\frac{f(t)}{t}-1)-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3}))p$ $<((\frac{f(t)}{t}-1)-(1+d_1+\frac{1}{d_3})(d_2+\frac{1}{d_3}))(f(p)+a)$ $<t$ , contradiction. a little ridiculous_ (´_`」 ∠) _