Prove inequality with logarithms:$\log_{a}{\frac{a+b}{2}} + \log_{b}{\frac{a+b}{2}} \ge 2$ Let $ a, b \in (1, \infty)$. Prove that:
$$\log_{a}{(\frac{a+b}{2})} + \log_{b}{(\frac{a+b}{2})} \ge 2$$
I tried switching the bases on each of the logaritm but I got stuck:
$$\frac{\log_{\frac{a+b}{2}}{(ab)}}{\log_{\frac{a+b}{2}}{(a)}\log_{\frac{a+b}{2}}{(b)}}$$

Let's set $f(x)=\dfrac 1{\ln(x)}$
We have $f''(x)=\dfrac{2+\ln(x)}{x^2\ln(x)^3}>0$ on $(1,+\infty)$, thus $f$ is convex.
The result is just an application of the convexity inequality
$$f\left(\frac{a+b}2\right)\le \frac 12f(a)+\frac 12f(b)\iff \frac 1{\ln(a)}+\frac 1{\ln(b)}\ge \frac 2{\ln(\frac{a+b}2)}$$
