Yan and Jacob play the following game. Yan shows Jacob a weighted 4-sided die labelled 1, 2, 3, 4, with weights $\frac{1}{2}, \frac{1}{3}, \frac{1}{7}, \frac{1}{42}$, respectively. Then, Jacob specifies 4 positive real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ such that $x_{1}+\cdots+x_{4}=1$. Finally, Yan rolls the dice, and Jacob earns $10+\log \left(x_{k}\right)$ dollars if the die shows $k$ (note this may be negative). Which $x_{i}$ should Jacob pick to maximize his expected payoff?

(Here log is the natural logarithm, which has base $e \approx 2.718$.)
Jacob should pick $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(\frac{1}{2}, \frac{1}{3}, \frac{1}{7}, \frac{1}{42}\right)$. More generally, suppose the weights are $p_{1}, \ldots, p_{4}$. Then Jacob's expected payoff is

$$
10+\sum_{i=1}^{4} p_{i} \log \left(x_{i}\right)=10+\sum_{i=1}^{4} p_{i} \log p_{i}+\sum_{k=1}^{4} p_{i} \log \left(\frac{x_{i}}{p_{i}}\right)
$$

Now, by JENSEN's INEQUALITY on the concave function $\log x$, we obtain

$$
\sum_{i=1}^{4} p_{i} \log \left(\frac{x_{i}}{p_{i}}\right) \cdot \leq \log \left(\sum_{i=1}^{4} p_{i} \cdot \frac{x_{i}}{p_{i}}\right)=\log 1=0
$$

and equality occurs exactly when $\frac{x_{1}}{p_{1}}=\frac{x_{2}}{p_{2}}=\frac{x_{3}}{p_{3}}=\frac{x_{4}}{p_{4}}$; that is, when $x_{i}=p_{i}$ for every $i$.