<s>[INST] 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:
$$\frac{\log_{\frac{a+b}{2}}{(ab)}}{\log_{\frac{a+b}{2}}{(a)}\log_{\frac{a+b}{2}}{(b)}}$$
 [/INST] We have to prove that
$$\frac{\ln\left(\frac{a+b}{2}\right)}{\ln(a)}+\frac{\ln\left(\frac{a+b}{2}\right)}{\ln(b)}\geq 2$$ this is $$\ln\left(\frac{a+b}{2}\right)(\ln(a)+\ln(b))\geq 2\ln(a)\ln(b)$$ or
$$\ln\left(\frac{a+b}{2}\right)\ln(ab)\geq 2\ln(a)\ln(b)$$ Now we have by AM-GM:
$$\frac{a+b}{2}\geq \sqrt{ab}$$
taking the logarithm on both sides
$$\ln\left(\frac{a+b}{2}\right)\geq \frac{1}{2}\ln(ab)$$ now is
$$\ln\left(\frac{a+b}{2}\right)\ln(ab)\geq \frac{1}{2}(\ln(ab))^2$$
Now we have to show that $$(\ln(ab))^2\geq 4\ln(a)\ln(b)$$ this is true since
$$(\ln(a)+\ln(b))^2\geq 4\ln(a)\ln(b)$$
this is equivalent to
$$(\ln(a)-\ln(b))^2\geq 0$$
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