"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 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:
$$
\sum_i \tilde\rho_{ii} = \sum_i \sum_n p_n a_{ni}a_{ni}^*
=\sum_n p_n \delta_{nn} = \sum_n p_n = 1
$$
