"okay but I don't know how I can apply this inequality to a real case. I edited an example to my post" (OP in a comment to a different answer)
$$ \rho²=\begin{pmatrix}\rho_{11}&\rho_{12}\\\rho_{21}&\rho_{22}\end{pmatrix}^2=\begin{pmatrix}\rho_{11}^2+\rho_{12}\rho_{21}&\rho_{11}\rho_{12}+\rho_{12}\rho_{22}\\\rho_{21}\rho_{11}+\rho_{22}\rho_{21}&\rho_{21}\rho_{12}+\rho_{22}^2\end{pmatrix} $$
By the spectral theorem, every density operator can be written in some basis as: $$ \hat\rho = \sum_{n} p_n|\phi_n\rangle\langle\phi_n|\;. $$ That is, a basis can be chosen such that $\rho$ is diagonal. And we have $$ \sum_n p_n= 1 $$
For your "real case" example, the matrix representation of the density operator in this basisin the basis where it is diagonal is: $$ \rho =\begin{pmatrix}p_{1}&0\\0&p_{2}\end{pmatrix} $$ and $Tr(\rho)$ is clearly equal to 1.
$$ Tr(\rho^2) = \sum_n p_n^2\;, $$ which is clearly less than or equal to 1.
Update (to address comments):
Suppose that the density matrix is re-written in a different basis $|\tilde \phi_n\rangle$, where it does not appear manifestly diagonal: $$ \hat \rho = \sum_{n}p_n|\phi_n\rangle\langle\phi_n| =\sum_{nij}p_n a_{ni}a_{nj}^*|\tilde \phi_i\rangle\langle\tilde \phi_j| \equiv \sum_{ij}\tilde\rho_{ij}|\tilde \phi_i\rangle\langle\tilde \phi_j|\;, $$ where $$ \tilde\rho_{ij} = \sum_n p_n a_{ni}a_{nj}^*\;, $$ and where, we also know that the transformation matrix is unitary. I.e., $$ \sum_i a_{ni}a_{mi}^* = \delta_{nm}\;, $$ which we know to be true since we require both bases to be orthonormal.
So then, let's compute the trace in the $\tilde \phi$ basis (we could use any basis really). The trace is: $$ Tr(\rho) = \sum_{k}\sum_{ij}\tilde\rho_{ij}\langle\tilde\phi_k|\tilde \phi_i\rangle\langle\tilde \phi_j|\tilde \phi_k\rangle =\sum_i \tilde \rho_{ii}\;, $$ which is what we expect.
But now recall the above definition of $\tilde \rho_{ij}$ to see that: $$ \sum_i \tilde\rho_{ii} = \sum_i \sum_n p_n a_{ni}a_{ni}^* =\sum_n p_n \delta_{nn} = \sum_n p_n = 1 $$