A conjecture about traces of projections

I prove only $y\ge \frac x2(3x-1)$. We know that $A+B+C\ge 0$, so let $\lambda_1,\dots,\lambda_n$ be its eigenvalues. Then $$ \frac 1n Tr(A+B+C)=\frac 1n \sum \lambda_i=\bar \lambda= 3x $$ and $$ Tr((A+B+C)^2)=\sum \lambda_i^2\ge n \bar \lambda^2=9nx^2. $$ But $$ (A+B+C)^2=A^2+B^2+C^2+AB+AC+BA+BC+CA+CB $$ and so $$ Tr((A+B+C)^2)=Tr(A+B+C)+2Tr(AB+BC+CA)=3nx+6ny. $$ Hence $3nx+6ny\ge 9nx^2$ which yields $y\ge \frac x2(3x-1)$.

${\bf{Edit:}}$

The bound given by the OP is attained for every $n$ and every admissible $x$: Set $$ A_0=\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix},\quad B_0=\begin{pmatrix} 1/4&-\sqrt{3}/4\\ -\sqrt{3}/4&3/4 \end{pmatrix}\quad\text{and}\quad C_0=\begin{pmatrix} 1/4&\sqrt{3}/4\\ \sqrt{3}/4&3/4 \end{pmatrix}. $$ Then $Tr(A_0B_0+B_0C_0+C_0A_0)=\frac 34$.

Let now $n$ be given, and $x=\frac {n+k}{3n}$ be admissible, i.e., $0\le k\le n/2$. Define $A$ to have $k$ times the block matrix $A_0$ on the diagonal, and complete the diagonal with $n-2k$ times 1: $$ A=\begin{pmatrix} A_0&&&&& \\&\ddots&&& 0& \\ && A_0&&&\\ &&&1 && \\ &0&&& \ddots &\\ &&&&& 1\end{pmatrix}. $$ Define $B$ to have $k$ times the block matrix $B_0$ on the diagonal, and complete the diagonal with $n-2k$ times 0: $$ B=\begin{pmatrix} B_0&&&&& \\&\ddots&&& 0& \\ && B_0&&&\\ &&&0 && \\ &0&&& \ddots &\\ &&&&& 0\end{pmatrix}. $$ Define $C$ to have $k$ times the block matrix $C_0$ on the diagonal, and complete the diagonal with $n-2k$ times 0: $$ C=\begin{pmatrix} C_0&&&&& \\&\ddots&&& 0& \\ && C_0&&&\\ &&&0 && \\ &0&&& \ddots &\\ &&&&& 0\end{pmatrix}. $$

Then $x=\frac{1}{3n}Tr(A+B+C)=\frac{n+k}{3n}$ (which implies $\frac kn=3x-1$) and $$ y=\frac{1}{3n}Tr(AB+BC+CA)=\frac{1}{3n}Tr(k(A_0B_0+B_0C_0+C_0A_0))=\frac{1}{3n}\frac 34k= \frac 14 \frac kn=\frac{3x-1}{4}, $$ hence the bound is attained.