Asian Option pricing payoff inequality

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I am trying to prove the following inequality:

$$ \left( \exp \left(\frac{1}{n} \sum_{i=1}^n \log(S_{t_i}) \right) - K \right)^+ \le \left( \left( \frac{1}{n} \sum_{i=1}^n S_{t_i} \right) - K \right)^+ $$

I can observe the inequality when I use fixed numbers, but I would like to establish it for any set of positive numbers.

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You need the AM-GM inequality: $$\sqrt[n]{x_1 x_2 \cdots x_n} \leq \frac{x_1 +x_2 + \cdots + x_n}{n} $$ (see, for example:https://artofproblemsolving.com/wiki/index.php/Arithmetic_Mean-Geometric_Mean_Inequality )

Then you have for the exponential term (since nothing happens with the $K$ term or the positive-part term I'll ignore them): $$ \begin{eqnarray} \exp \left( \frac{1}{n}\sum_{i=1}^n \log (S_{t_i}) \right) &=& \exp^{\log S_{t_1}^{1/n}} \cdot \exp^{\log S_{t_2}^{1/n}} \cdots \exp^{\log S_{t_n}^{1/n}} \\ & = & S_{t_1}^{1/n} \cdot S_{t_2}^{1/n} \cdots S_{t_n}^{1/n} \\ & \leq & \frac{\sum_{i=1}^n S_{t_i}}{n} \end{eqnarray}$$