PomeLemma 8.2.6.473C52E484B0 If $0<f_0\leq f\leq f_1$, then
\[ Q:=\sum _ iF_ i\lambda _ i^2 \geq \frac{f_0}{m}S-\frac{n-m}{m}f_1^2, \qquad \mathcal F\geq 1. \]
Lemma 8.2.6473C52E484B0
Proof. The derivative of the normalized root is $F_ i=fS_{m-1}(\lambda |i)/(mS_ m)$. The splitting identity gives
\[ Q=\frac fm S-\frac{(m+1)f}{mS_ m}S_{m+1}. \]
For $m=n$, set $S_{n+1}=0$. If $S_{m+1}\leq 0$, the second term is nonnegative. Otherwise $\lambda \in \Gamma _{m+1}$ and Maclaurin gives $S_{m+1}\leq \binom n{m+1}f^{m+1}$. Since $S_ m=\binom nm f^ m$, the subtracted term is at most $(n-m)f^2/m$. This proves the first assertion. Summing the derivatives gives $\mathcal F=f\sigma _{m-1}/\sigma _ m\geq 1$, again by Maclaurin. $\square$
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