How do I prove the following, where $f:[0,\infty)\to[0,\infty)$:
For $x>0$ if $f(x)$ satisfies $$2f(x)f(1)\leq f(x)f(1)+1\leq 2f(2x)$$ then $f(x)$ is a constant function.
I have out found out that when $f(x)=1$ this inequality is satisfied
So I tried to prove by contradictory by supposing where when there exists $a$ that satisfies $f(a)>1$ or when satisfies $0<f(a)<1$, but I failed.
Intro: Let $f:[0,\infty]\to[0,\infty]$ such that: $$2f(x)f(1)\stackrel{(A)}{\le}f(x)f(1)+1\stackrel{(B)}{\le}2f(2x) $$
Lower bound: I'll now create a recurrence sequence $(a_n)_\mathbb N$ such that $f(x)\ge a_n\; \forall n$, let $a_1=0$: $$2f(x)\stackrel{(B)}{\ge}f(x/2)f(1)+1\ge 0+1 \implies f(x)\ge 1/2=:a_2 \;\forall x\in[0,\infty]. $$ Repeating the same steps with $a_2$: $$2f(x)\stackrel{(B)}{\ge}f(x/2)f(1)+1\ge a_2^2+1 \implies f(x)\ge \frac{a_2^2+1}{2}=:a_3\;\forall x\in[0,\infty].$$ So we get that $f(x)\ge a_n$ with $a_n$ that meets $a_{n+1}=\frac{a_n^2+1}{2}$. Furthermore $a_n$ is monotone ($\frac{a_n^2+1}{2}\ge a_n \iff \frac{(a_n-1)^2}{2}\ge 0$) and its limit to infinity is clearly $1$: $$l=\lim a_n=\lim \frac{a_n^2+1}{2}=\frac{l^2+1}{2}\implies l=1.$$ Hence $f(x)\ge1\; \forall x\in[0,\infty].$
Upper bound: From (A) we get $f(x)\le f(1)^{-1}$ but for the previous part $f(1)^{-1}\le 1\implies f(x)\le 1\; \forall x\in[0,\infty].$
Conclusion: $1\le f(x)\le 1\implies f$ is the constant function $f(x)=1.$
