Let $M \in S _ { n } ^ { + } ( \mathbf { R } )$ be a non-zero matrix. In the rest of this part, you may use without proof the inequality below $\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \left( \mathbf { R } _ { + } \right) ^ { n }$, $$2 \max \left\{ x _ { 1 } , \ldots , x _ { n } \right\} \left( \frac { 1 } { n } \sum _ { k = 1 } ^ { n } x _ { k } - \prod _ { k = 1 } ^ { n } x _ { k } ^ { 1 / n } \right) \geq \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \left( x _ { k } - \prod _ { j = 1 } ^ { n } x _ { j } ^ { 1 / n } \right) ^ { 2 }$$ Deduce that $$\frac { \operatorname { Tr } ( M ) } { n } - \operatorname { det } ^ { 1 / n } ( M ) \geq \frac { \left\| M - \operatorname { det } ^ { 1 / n } ( M ) I _ { n } \right\| _ { 2 } ^ { 2 } } { 2 n \| M \| _ { 2 } }$$
Let $M \in S _ { n } ^ { + } ( \mathbf { R } )$ be a non-zero matrix. In the rest of this part, you may use without proof the inequality below $\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \left( \mathbf { R } _ { + } \right) ^ { n }$,
$$2 \max \left\{ x _ { 1 } , \ldots , x _ { n } \right\} \left( \frac { 1 } { n } \sum _ { k = 1 } ^ { n } x _ { k } - \prod _ { k = 1 } ^ { n } x _ { k } ^ { 1 / n } \right) \geq \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \left( x _ { k } - \prod _ { j = 1 } ^ { n } x _ { j } ^ { 1 / n } \right) ^ { 2 }$$
Deduce that
$$\frac { \operatorname { Tr } ( M ) } { n } - \operatorname { det } ^ { 1 / n } ( M ) \geq \frac { \left\| M - \operatorname { det } ^ { 1 / n } ( M ) I _ { n } \right\| _ { 2 } ^ { 2 } } { 2 n \| M \| _ { 2 } }$$