A conjecture generalization of Karamata inequality
Conditions (1,2,3) are not enough for the inequality to hold. Take $n=3$, $x_1=x_2=3,x_3=0$, $y_1=3,y_2=y_3=0$. Then we need the multiset $(3,3,0,1,1,1)$ (all three $x$'s and 3 times mean of $y$'s) to majorate $(3,0,0,2,2,2)$. But four largest elements of the first multiset have sum $3+3+1+1=8$, while in the second it is $3+2+2+2=9$. So, the claim does not hold in full generality. To be more explicit, we get an opposite inequality for $f(x)=\max(x-2,0)$.
Conditions (1) and
(2') $x_1+...+x_i \geqslant y_1+.....+y_i$ and $x_{i+1}+...+x_n \leqslant y_{i+1}+...+y_n$ for $i=1,....,n-1$
are enough. We prove it by verifying that the multiset of $2n$ numbers $A:=\{x_1,\dots,x_n,y,y,\dots,y\}$ majorates the multiset $B:=\{y_1,\dots,y_n,x,x,\dots,x\}$, where $x=\frac1n \sum x_i$, $y=\frac1n \sum y_i$. Without loss of generality $x\leqslant y$, else change signs of all $x$'s and $y$'s. We have to check that the sum $w_m$ of $m$ largest elements of $B$ does not exceed the sum of $m$ largest elements of $A$. Let $m$ largest elements of $B$ be $y_1,\dots,y_s$ and $(m-s)$ times $x$. Consider two cases.
1-st case. $s\leqslant n-1$. Then $w_m=y_1+\dots+y_s+(m-s)x\leqslant x_1+\dots+x_s+(m-s)y$.
2-nd case. $s=n$. Then $w_m=y_1+\dots+y_n+(m-n)x\leqslant n\cdot y+x_1+\dots+x_{m-n}$.
In both cases we found $m$ elements of $A$ with a sum at least $w_m$, as desired.