# Find $f:(1,\infty) \to \mathbb{R}$

Find $f:(1,\infty) \to \mathbb{R}$, where \begin{align*} &f(x) \leq \frac{x-2}{\ln 2}, \: \forall x>1 \\ &f(x^3+1) \leq 3f(x+1), \: \forall x>0 \\ &f(x)+f\left(\frac{x}{x-1} \right) \geq 0, \: \forall x>1 \end{align*}

All I got, which is also obvious, is the fact that $f(2)=0$. I tried to at least guess the function, without success. Also, substituting $x \to \frac{x}{x-1}$ in the last inequality makes it unchanged, thus this doesn't help either... My last thought was that the $3$ in the second inequality getting out of the function must somehow be related to $\log$s.

As a side note, I found this as a real analysis/calculus problem.

• Identically zero – jnyan Nov 25 '17 at 18:30

You are on the right track! Let $g(x):=f(x+1)$ then the second and the third inequalities say that for all $x>0$, $$g(x^3) \leq 3g(x), \quad g(x)\geq -g\left(1/x \right).$$ Maybe $g$ is a logaritmic function... Let's try with $g(x)=\log_b(x)$, that is $f(x)=\log_b(x-1)$. For such $f$, the second and the third inequalities are equalities.
The first inequality says that the graph of the concave function $f$ (assume that $b>1$) is under the graph of its tangent line at $2$, i.e. $y=(x-2)/\ln(2)$. The tangent line is $$y=f(2)+f'(2)(x-2)=0+\frac{(x-2)}{\ln(b)}$$ Then we choose $b=2$ and $f(x)=\log_2(x-1)$ satisfies all the given inequalities.