# Inequality involving an increasing convex function

I am trying to prove/disprove the following statement:

Let $x_1 \geq \cdots \geq x_n$, $y_1 \geq \cdots \geq y_n$ be real numbers satisfying $x_1 + \cdots + x_k \leq y_1 + \cdots + y_k$ for all $1 \leq k \leq n$. Let $f:\mathbb{R} \to \mathbb{R}$ be an increasing convex function. Then one has $f(x_1) + \cdots + f(x_n) \leq f(y_1) + \cdots + f(y_n)$.

The case $n=1$ is trivial, and the case $n=2$ can be proved as following:

Assume $x_2 > y_2$, since otherwise the conclusion follows trivially. Then we can take $c_1 \geq c_2 \geq 0$ s.t. $\begin{cases} f(y_1)-f(x_1) \geq c_1 (y_1-x_1)\\ f(x_2)-f(y_2) \leq c_2(x_2-y_2) \end{cases}$

(for instance, if $f$ is differentiable, then we can take $c_1 = f'(x_1) \geq f'(x_2) = c_2$.)

Then we see that $f(y_1)+ f(y_2)-f(x_1)-f(x_2) \geq c_1(y_1-x_1) + c_2(y_2-x_2) \geq c_2(y_1+y_2-x_1-x_2) \geq 0$.

But I have difficulty extending this proof to the case $n\geq 3$. Any help/comment will be appreciated. Thanks.

Let $x_1+x_2+...+x_n=y_1+y_2+...+y_{n-1}+y_n'.$
Thus, $y_n'\leq y_n$, $(y_1,y_2,...,y_n')\succ(x_1,x_2,...,x_n)$ and by Karamata we obtain: $$f(x_1)+f(x_2)+...+f(x_n)\leq f(y_1)+f(y_2)+...+f(y_n')\leq f(y_1)+f(y_2)+...+f(y_n).$$