How to prove that $\mathrm{tr}(\rho^2)=1$ if and only if the state is pure? 
How can I prove that $\mathrm{tr}(\rho^2) $ = 1 if and only if the state is pure? 

My idea: 
I know how to show that $\mathrm{tr}(\rho^2) \leq 1$ and from there I am trying to show by contradiction that $\mathrm{tr}(\rho^2) = 1$ can only but true for pure state, but I am kind of stuck. 
I just need a hint on how to prove this. 
 A: Since an arbitrary $\rho$ is self-adjoint, it has the spectral decomposition $\rho = \sum_n \rho_n |\psi_n><\psi_n|$, in terms of an orthonormal basis $\{ |\psi_n>\}$, which here we pick discrete for simplicity.
Hermiticity implies $\rho_n = \rho^*_n$. $\mathrm{Tr} \rho = 1$ implies $\sum_n \rho_n =1$. Semi-positivity implies $0 \leq \rho_n$. Together they imply $0 \leq \rho_n \leq 1$, which implies $\rho^2_n \leq \rho_n$. Hence, $\mathrm{Tr} \rho^2 = \sum_n \rho^2_n \leq \sum_n \rho_n = 1$ and so $\mathrm{Tr} \rho^2 \leq 1$ for a generic state, as you mention.
Now let's start assuming that  $\mathrm{Tr} \rho^2 =1$. Following the inequalities we just wrote, this implies that  $\rho^2_n = \rho_n$ for all $n$ i.e. $\rho^2 = \rho$. In particular this implies that $\rho_n = 1$ or $\rho_n = 0$. More precisely, due to the trace condition $\mathrm{Tr} \rho = 1$, only one $\rho_n$ is equal to one while the others vanish. This is a pure state.
The reverse implication is direct.
A: If you diagonalize the density matrix into eigenvectors $|\psi_i\rangle$ and eigenvalues (probabilities) $\lambda_i\in\mathbb{R}_+$, then the conditions $\text{tr}(\rho)=\text{tr}(\rho^2)=1$ give
$$\sum_{i}\lambda_i=1$$
and
$$\sum_{i}\lambda_i^2=1,$$
respectively. Showing that these equations are inconsistent unless only one of the $\lambda_i$ is nonzero is equivalent to showing that $\text{tr}(\rho^2)=1$ implies that the system is in a pure state.
Hint: Square the first condition and subtract away the second to reach a contradiction.
I hope this helps!
PS: Another way to rephrase this is to show that given any positive-definite operator $A$, then the inequality $\langle A\rangle^2-\langle A^2\rangle\leq 0$ is saturated if and only if $A=|\psi\rangle\langle\psi|$ for some vector $\psi\in\mathcal{H}$.
