A conjecture about traces of projections Let $M_n$ denote the space of all $n\times n$ complex matrices. Define $\tau:M_n\rightarrow \mathbb{C}$ by $$\tau(X)=\frac{1}{n}\sum_{i=1}^n x_{ii},$$ where of course $X=[x_{ij}]\in M_n$. Recall that a matrix $P\in M_n$ is called an "orthogonal projection" if $P=P^*=P^2$. Let $A, B, C$ be orthogonal projections in $M_n$ and define two quantities - $$x=\frac{1}{3}\tau(A+B+C)$$ and $$y=\frac{1}{3}\tau(AB+BC+CA).$$ 
It is easy to see that for any orthogonal projections $A,B,C\in M_n$ the value of $x$ lies in $[0,1]$. I want to investigate the case when $x\in [\frac{1}{3},\frac{1}{2}]$. My goal is to compute the infimum of $y$ under this constrain on $x$.
My observation is that the minimum value of $y$ is $\frac{3x-1}{4}$ when $x\in [\frac{1}{3},\frac{1}{2}]$ and for all $n\in \mathbb{N}$.
However I am unable to prove this and I wondered if people here may be able to see why this might hold, or provide a counterexample to this conjecture.

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.
