An m-qubit state, inferring some inequality in QIT

[billtodd]

Homework Statement

Let $\sigma$ be an $m$-qubit such that $Trace(\sigma^2)\le \frac{1+\epsilon^2}{2^m}$, then $D(\sigma , I/2^m)\le \epsilon$

Relevant Equations

$D(\sigma,\rho)=1/2 Trace(|\sigma-\rho|)$ where $\rho,\sigma$ are quantum states; and there's an inequality between this and the fidelity, i.e. $D(\sigma,\rho)\le \sqrt{1-F(\sigma,\rho)^2}$, where $F(\sigma ,\rho)=Trace(\sqrt{\sigma^{1/2}\rho \sigma^{1/2}})$.

I am not sure how to show this inequality, I guess what is the relation between $Trace(\sigma^2)$ and $Trace(\sigma)$?
Here's my latest attempt, let me know how to proceed?
$$D(\sigma, I/2^m)\le \sqrt{1-F(\sigma,I/2^m)^2}=\sqrt{1-(Tr(\sqrt{\sigma }))^2/2^m}$$