lower bounding logarithm of sums

Question

Is it true that

$$ \log\left(\sum_{i=1}^{n} \alpha_i\right) = \log\left(n \frac{1}{n}\sum_{i=1}^{n}\alpha_i\right) = \log(n) + \log\left(\frac{1}{n}\sum_{i=1}^{n} \alpha_i\right) \\ \geq \log(n) + \frac{1}{n}\sum_{i=1}^{n}\log(\alpha_i)$$ (where Jensen's inequality was applied in the last step)?

Answer 1

Yes, it is true. Jensen's inequality was applied in the last step $$\log \left( \frac{1}{n} \sum \limits_{i=1}^{n}{\alpha_i} \right) \geq \frac{1}{n} \sum _{i=1}^n {\log(\alpha_i)},$$ since $\log$ is concave.