Does large quantum Fisher information imply Bell correlations?

DOI: 10.1103/PhysRevA.99.040101
Eq(8):

Try to use Lagerange's multipliers method, only find it too complicated:

\[\begin{cases} \sum_{n=1}{\frac{p_nn^2}{1+p_0}}\\ \sum_{n=1}{p_nn}=\left( 1+p_0 \right) \bar{n}\\ \sum_{n=0}{p_n}=1\\ \end{cases} \\ \begin{cases} \frac{n^2}{1+p_0}=\alpha n+\beta\\ \sum_{n=1}{p_nn}=\left( 1+p_0 \right) \bar{n}\\ \sum_{n=0}{p_n}=1\\ \end{cases} \\ \begin{cases} \frac{p_1+4p_2}{1+p_0}\\ p_1+2p_2-\left( 1+p_0 \right) \bar{n}=0\\ p_0+p_1+p_2=1\\ p_0-a^2=0\\ p_1-b^2=0\\ p_2-c^2=0\\ \end{cases} \\ \begin{cases} -\frac{p_1+4p_2}{\left( 1+p_0 \right) ^2}=-\lambda _1\bar{n}+\lambda _2+\lambda _3\\ \frac{1}{1+p_0}=\lambda _1+\lambda _2+\lambda _4\\ \frac{4}{1+p_0}=2\lambda _1+\lambda _2+\lambda _5\\ 0=-2a\lambda _3\\ 0=-2b\lambda _4\\ 0=-2c\lambda _5\\ p_1+2p_2-\left( 1+p_0 \right) \bar{n}=0\\ p_0+p_1+p_2=1\\ p_0-a^2=0\\ p_1-b^2=0\\ p_2-c^2=0\\ \end{cases} \\ \begin{cases} -\frac{p_1+4p_2}{\left( 1+p_0 \right) ^2}=-\lambda _1\bar{n}+\lambda _2\\ \frac{1}{1+p_0}=\lambda _1+\lambda _2\\ \frac{4}{1+p_0}=2\lambda _1+\lambda _2\\ p_1=0\\ p_1+2p_2-\left( 1+p_0 \right) \bar{n}=0\\ p_0+p_1+p_2=1\\ p_0-a^2=0\\ p_1-b^2=0\\ p_2-c^2=0\\ \end{cases} \]