We know that $4f(x)-3x\ge0$, so, we have $4f(4f(x)-3x)-3(4f(x)-3x)\ge0$ Then: $4x-3(4f(x)-3x)\ge0$ $x\ge f(x)$ Using this: $4f(x)-3x\ge f(4f(x)-3x)$ $4f(x)-3x\ge x$ $f(x)\ge x$ So $x \ge f(x) \ge x$ and $f(x)=x$

You've got a mistake, rsa: rsa365 wrote: We know that $4f(x)-3x\ge0$, so, we have $4f(4f(x)-3x)-3(4f(x)-3x)\ge0$ Then: $4x-3(4f(x)-3x)\ge0$ $\frac{13}{12} x \ge f(x)$ Let $a_1=\frac{3}{4}$ and $b_1 = \frac{13}{12}$. For $n>1$, let $a_n = \frac{3b_{n-1}+1}{4b_{n-1}}$ and $b_n = \frac{3a_n+1}{4a_n}$. We can show that $a_nx \leq f(x) \leq b_nx$ for every $n \in \mathbb{N}$. We can prove inductively that $\left\{a_n\right\}$ is increasing, that $\left\{b_n\right\}$ is decreasing, and that $a_n < 1 < b_n$ for every $n$. Finally, if $n>1$, \begin{align*} 0<1-a_n &= 1-\left(\frac{3}{4}+\frac{1}{4b_n}\right) = \frac{1}{4}-\frac{a_{n-1}}{1+3a_{n+1}} \\ &=\frac{1-a_{n-1}}{4\left(1+3a_{n-1}\right)} \leq \frac{1-a_{n-1}}{4\left(1+3a_1\right)}=\frac{1-a_{n-1}}{13}\,. \end{align*} Hence, $0 < 1-a_n \leq \frac{1}{4\cdot13^{n-1}}$ for every $n$. As $n \to \infty$, $a_n \to 1$, so $b_n \to 1$ as well. This proves that $f$ is the identity.

g(x)=4f(x)-3x we have:g(g(x))+3g(x)=4x let:$x_{1}=x $ $x_{i+1}=f(x_{i})$ we have that $ x_{i+2}+3x_{i+1}=4x_{i}$and now it is very easy to see that $ x_{i} $ is constant so g(x)=x and f(x)=x

a similar problem is : f a function from positives numbers to positives numbers such that f(f(x)-x)=6x find f this problem is from an romania shortlit for the national olympiad from 2008

There's something missing in this problem statement. Below is the accurate translation (originally from the Turkish source): Find all $f:[0,\infty)\to [0,\infty)$, satisfying, $\bullet$ $4f(x)\geq 3x$ $\bullet$ $f(4f(x)-3x)=x$ $\bullet$ $(f(x)+x)f(f(x))\leq 2xf(x)$, for every $x\in [0,\infty)$. With this correction, here is a solution. The first item above is there to make sure that, we can safely insert $4f(x)-3x$ into function's argument. Take the third condition, and study it with $x\mapsto 4f(x)-3x$ substitution, to arrive at: $$ (f(4f(x)-3x)+x)f(f(4f(x)-3x))\leq 2(4f(x)-3x)f(4f(x)-3x) \implies 2xf(x)\leq 2f(x)(4f(x)-3x), $$or equivalenly, $f(x)\geq x$ pointwise. Next, use the second item, to note that, $x=f(4f(x)-3x)\geq 4f(x)-3x \implies x\geq f(x)$ pointwise. Hence, $f(x)=x$ is the only solution.

zamfiratorul wrote: a similar problem is : f a function from positives numbers to positives numbers such that f(f(x)-x)=6x find f this problem is from an romania shortlit for the national olympiad from 2008 similar problem appeared in India TST 2001/3 by replacing 6 with 2.