Let $\hat { \lambda } = \left( \lambda _ { 1 } \geqslant \cdots \geqslant \lambda _ { n + 1 } \right) \in \mathbb { R } ^ { n + 1 }$ and $\widehat { \mu } = \left( \mu _ { 1 } \geqslant \cdots \geqslant \mu _ { n } \right) \in \mathbb { R } ^ { n }$. Let $x \in \mathbb { R }$. Form $$\widehat { \lambda } ^ { \prime } = \left( \lambda _ { 1 } \geqslant \cdots \geqslant \lambda _ { i } \geqslant x > \lambda _ { i + 1 } \geqslant \cdots \geqslant \lambda _ { n + 1 } \right)$$ by choosing the integer $i \in \{ 0 , \ldots , n + 1 \}$ appropriately. If $x > \lambda _ { 1 }$, we thus have $i = 0$, while if $x \leqslant \lambda _ { n + 1 }$, we have $i = n + 1$. Similarly form $$\widehat { \mu } ^ { \prime } = \left( \mu _ { 1 } \geqslant \cdots \geqslant \mu _ { j } \geqslant x > \mu _ { j + 1 } \geqslant \cdots \geqslant \mu _ { n } \right) .$$ Assume that $\widehat { \lambda }$ and $\widehat { \mu }$ are interlaced. Show that $j \leqslant i \leqslant j + 1$. By examining each of the two cases $j = i$ or $i - 1$, show that $\widehat { \lambda } ^ { \prime }$ and $\widehat { \mu } ^ { \prime }$ are interlaced.