If $f:[0,\infty]\rightarrow [0,\infty]$ is such that for all $x\in [0,\infty]$ :$${(f(x))}^2=1+xf(x+1)$$ Prove that for all $x\ge 1$ $$\frac{x+1}{2}\le f(x)\le 2(x+1)$$
First of all I have never had experience proving such inequalities and since nothing is said about the differentiability I don't know whether calculus would be a good approach. We can however draw the following conclusions for all $x\ge 1$
- As $f(x)\ge 1$ we have $$f(x)^2=1+xf(x+1)\ge 1+x \implies f(x)\ge \sqrt{x+1}$$
- By Bernoulli's inequality $$f(x)=\sqrt{1+xf(x+1)}\le 1+\frac{xf(x+1)}{2}$$
- As $f(x+1)\le f^2(x+1)$ and $x\le x^2$ $$f(x)^2=1+xf(x+1)\le 1+x^2f^2(x+1)$$ $$(f(x)-xf(x+1))(f(x)+xf(x+1))\le 1$$
That's it, I couldn't get any more ideas ...
P.S. The problem is from here.
A stronger lower bound
Define the functions $a_n: [0, \infty) \to \Bbb R$ recursively as $$ \begin{align} a_0(x) &= 1 \, ,\\ a_{n+1}(x) &= \sqrt{1+xa_n(x+1)} \, . \end{align} $$ In Nested Radical of Ramanujan it is shown that for all $x \ge 0$ $$ \lim_{n\to \infty} a_n(x) = x+1 \, . $$ On the other hand, $f(x) \ge a_n(x)$ follows from the given functional equation with induction. Therefore $$ f(x) \ge x + 1 \text{ for all } x \ge 0. $$
There is no upper bound
Choose an arbitrary function $g:[0, 1] \to \Bbb R$ with $g(0) = 1$ and $g(x) \ge 1+x$. Define $f$ on $[0, 1]$ as $f(x) = g(x)$, and extend the definition to all intervals $(n, n+1]$, $n = 1, 2, 3, \ldots$, with $$ f(x) = \frac{f(x-1)^2 - 1}{x-1} \, . $$ Note that $f(x-1) \ge x$ on $(n-1, n]$ implies that $f(x) \ge x+1$ on $(n, n+1]$, so that $f$ is well-defined for all $x \ge 0$ and satisfies the functional equation.
But $f(1) = g(1)$ can be arbitrarily large.
Remark: If $g$ is chosen as a differentiable function with $g(1) = 2g(0)g'(0)$ then $f$ is continuous at $x=1$ and therefore continuous everywhere. An example would be $g(x) = 1+kx + (k-1)x^2$ for any $k \ge 1$.