Suppose that $Q_1,\ldots,Q_m\in M_n(\mathbb R)$ are positive definite, $v_1,\cdots,v_m\in \mathbb{R}^n$ are $m$ given vectors and $\alpha_1,\ldots,\alpha_m$ are $m$ nonnegative real numbers that sum to $1$. How to show that $$ 0\le \sum_{i=1}^{m}\alpha_i v_i^TQ_iv_i-\left(\sum_{i=1}^m\alpha_iQ_iv_i\right)^T\left(\sum_{i=1}^m\alpha_iQ_i\right)^{-1}\left(\sum_{i=1}^m\alpha_iQ_iv_i\right)\le 1? $$
Any hints will be appreciated.
The upper bound $1$ is definitely wrong. If the difference is in middle is nonzero, by scaling up each $v_i$, the upper bound $1$ will be violated.
The lower bound is correct. We may assume that each $\alpha_i$ is nonzero. By absorbing $\alpha_i$ into $Q_i$, we may rewrite the inequality as $$ 0\le \sum_{i=1}^{m}v_i^TQ_iv_i-\left(\sum_{i=1}^mQ_iv_i\right)^T\left(\sum_{i=1}^mQ_i\right)^{-1}\left(\sum_{i=1}^mQ_iv_i\right).\tag{1} $$ Now let $w_i=Qv_i$. Then the inequality can be further rewritten as $$ 0\le \sum_{i=1}^{m}w_i^TQ_i^{-1}w_i-\left(\sum_{i=1}^mw_i\right)^T\left(\sum_{i=1}^mQ_i\right)^{-1}\left(\sum_{i=1}^mw_i\right).\tag{2} $$ By mathematical induction, it suffices to prove the inequality for the case $m=2$ only. In other words, it suffices to show that $$ 0\le u^TP^{-1}u+v^TQ^{-1}v -(u+v)^T(P+Q)^{-1}(u+v)\tag{3} $$ for any positive definite $P,Q\in M_n(\mathbb R)$ and any $u,v\in\mathbb R^n$. To prove this last inequality, we use the matrix identities \begin{align} X^{-1} - (X+Y)^{-1} &= X^{-1} (X^{-1} + Y^{-1})^{-1} \color{red}{X}^{-1},\\ (X+Y)^{-1} &= X^{-1}(X^{-1}+Y^{-1})^{-1}\color{blue}{Y}^{-1}, \end{align} so that \begin{align} &u^TP^{-1}u + v^TQ^{-1}v - (u^T+v^T)(P+Q)^{-1}(u+v)\\ &=u^T [P^{-1} - (P+Q)^{-1}] u\\ &-u^T(P+Q)^{-1}v\\ &-v^T(P+Q)^{-1}u\\ &+v^T [Q^{-1} - (P+Q)^{-1}] v\\ &=u^TP^{-1} (P^{-1} + Q^{-1})^{-1} P^{-1}u\\ &-u^TP^{-1} (P^{-1} + Q^{-1})^{-1} Q^{-1}v\\ &-v^TQ^{-1} (P^{-1} + Q^{-1})^{-1} P^{-1}u\\ &+v^TQ^{-1} (P^{-1} + Q^{-1})^{-1} Q^{-1}v\\ &=(P^{-1}u-Q^{-1}v)^T (P^{-1} + Q^{-1})^{-1} (P^{-1}u-Q^{-1}v)\\ &\ge0. \end{align}