A Property of an Increasing Concave Function Increasing concave functions with $f(0)=0$ like $x^{1/2}$ and $\log(1+x)$ satisfy the property that
$(x+1)f\left(\frac{1}{x+1}\right) - x f\left(\frac{1}{x}\right)$ is decreasing for all $x>0$. I have been trying to prove that this is generally true, but my attempt has not been successful so far. I am wondering whether anyone could help me (or disprove it). Thank you.
 A: Let $f(x)$ be a function that is strictly increasing and either concave up or down for all $x > 0,$ i.e., $f'(x) > 0$ and either $f''(x) > 0$ or $f''(x) < 0$ for all $x > 0.$ Consider the function $$g(x) = (x + 1) f \biggl(\frac 1 {x + 1} \biggr) - x f \biggl(\frac 1 x \biggr).$$
By the Product Rule and the Chain Rule, we have that $$g'(x) = f \biggl(\frac 1 {x + 1} \biggr) -\frac 1 {x + 1} f' \biggl(\frac 1 {x + 1} \biggr) - f \biggl(\frac 1 x \biggr) + \frac 1 x f' \biggl(\frac 1 x \biggr).$$ Consequently, for $g(x)$ to be decreasing for $x > 0,$ we need $g'(x) < 0$ for $x > 0,$ i.e., $$f \biggl(\frac 1 {x + 1} \biggr) + \frac 1 x f' \biggl(\frac 1 x \biggr) < f \biggl(\frac 1 x \biggr) + \frac 1 {x + 1} f' \biggl(\frac 1 {x + 1} \biggr) \tag{1}$$ for all $x > 0.$ By hypothesis that $f(x)$ is increasing for all $x > 0,$ we have that $$f \biggl(\frac 1 {x + 1} \biggr) < f \biggl(\frac 1 x \biggr)$$ for all $x > 0$ so that a sufficient (but not necessary) condition for inequality (1) to hold is $$\frac 1 x f' \biggl(\frac 1 x \biggr) < \frac 1 {x + 1} f' \biggl(\frac 1 {x + 1} \biggr) \tag{2}$$ for all $x > 0.$ Observe that if $f(x)$ is concave up on $x > 0,$ i.e., $f'(x)$ is increasing for $x > 0,$ then inequality (2) does not hold. For in this case, for all $x > 0,$ we have that $$x f' \biggl(\frac 1 {x + 1} \biggr) < x f' \biggl(\frac 1 x \biggr) < (x + 1) f' \biggl(\frac 1 x \biggr).$$ Unfortunately, there could exist functions $f(x)$ that are strictly increasing and concave down for all $x > 0$ and yet do not satisfy the inequality (1) for all $x > 0,$ all though I cannot think of one.
