How to prove that the following function is convex? I want to prove convexity of the following function:
$$f(x) = log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)$$
for any fixed $a, b \in (0, 1)$ and:


*

*$x\in(0,1)$

*$x\in(1, \infty)$


I'm trying to solve it for a very long time, tried to investigate the sign of second derivative, but it leads to very ugly equation which I am unable to handle. Any hints are appreciated. Thanks!

Edit:
$f''(x)$ equals to:
$$-\frac{{\left(\frac{x^{x - 1} {\left(p^{y} - 1\right)} x}{p - 1} + \frac{{\left(p^{x} - 1\right)}
    x^{y - 1} y}{p - 1} - \frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{{\left(p -
    1\right)}^{2}}\right)}^{2}}{{\left(\frac{{\left(x^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} +
    1\right)}^{2} \log\left(x\right)} +\\
+ \frac{\frac{x^{x - 2} {\left(p^{y} - 1\right)} {\left(x - 1\right)} x}{p - 1} + \frac{2 \, p^{x -
    1} x^{y - 1} x y}{p - 1} + \frac{{\left(p^{x} - 1\right)} p^{y - 2} {\left(y - 1\right)} y}{p - 1} -
    \frac{2 \, x^{x - 1} {\left(p^{y} - 1\right)} x}{{\left(p - 1\right)}^{2}} - \frac{2 \, {\left(p^{x}
    - 1\right)} x^{y - 1} y}{{\left(p - 1\right)}^{2}} + \frac{2 \, {\left(p^{x} - 1\right)}
    {\left(x^{y} - 1\right)}}{{\left(p - 1\right)}^{3}}}{{\left(\frac{{\left(p^{x} - 1\right)}
    {\left(x^{y} - 1\right)}}{p - 1} + 1\right)} \log\left(p\right)} -\\
- \frac{2 \, {\left(\frac{x^{x - 1} {\left(p^{y} - 1\right)} x}{p - 1} + \frac{{\left(p^{x} -
    1\right)} x^{y - 1} y}{p - 1} - \frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{{\left(p
    - 1\right)}^{2}}\right)}}{{\left(\frac{{\left(x^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} +
    1\right)} x \log\left(p\right)^{2}} + \frac{\log\left(\frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)}{p^{2}
    \log\left(x\right)^{2}} + \frac{2 \, \log\left(\frac{{\left(p^{x} - 1\right)} {\left(p^{y} - 1\right)}}{p - 1} + 1\right)}{p^{2} \log\left(p\right)^{3}}
$$
Some elements of that equation are positive, some of them are negative - I do not see any way of how to prove that $f''(x) > 0$.
I have also tried to find counter example by testing the Jensen's inequality on randomly generated data, but everything seems OK - the function seems to be convex, but I am unable to prove it.

You can write
$$
f(x) = \log_x \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)
     = \frac{\ln \left(1 + \frac{(x^a-1)(x^b - 1)}{x-1}\right)}{\ln x},
$$
and prove that $f''(x) > 0$ where needed.
