Let $p_1,\ldots,p_n$ be numbers such that $0 < p_i < 1$ and $\sum p_i = 1$.
Let also $x_1,\ldots,x_n$ be numbers such that $x_i \ge -1$.
Define the function (for $0
$f(y) = \sum_{i=1}^n \log\left(1+x_iy\right)p_i.$
Now let $p_{n+1}$ be a new number such that $0 < p_{n+1} < 1$, and let $x_{n+1}$ be a new number such that $x_{n+1} > x_i$ for all $i=1,\ldots,n$. Define the new function (for $0
$f^*(y) = \sum_{i=1}^n \log\left(1+x_iy\right)\left(p_i - \frac{p_{n+1}}{n}\right) + \log\left(1+x_{n+1}y\right)p_{n+1}.$
I believe I have already managed to show, with just basic algebra and some calculus, that $f^*(y) > f(y)$ for all $y$.
Problem: However, can someone help me show also that the maximum value of $f^*(y)$ lies to the right of the maximum value of $f(y)$? (In other words, that the $y$ that maximizes $f^*$ is greater than the $y$ that maximizes $f$.)