An inequality about positive matrices 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.
 A: 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}
