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Is the trace distance upper bounded by the Euclidean distance?

Suppose we have two pure state $|\psi\rangle$ and $|\phi\rangle$.

I was wondering whether the statement: $\||\psi\rangle\!\langle\psi|- |\phi\rangle\!\langle\phi|\|_{\rm tr}$ is at most the Euclidean distance between $|\psi\rangle$ and $|\phi\rangle$. I know that under the qubit condition, the trace distance is exactly half of Euclidean distance.

If it is right, then what should the vector representation be corresponding to these two pure states?