Suppose $V$ is a finite dimensional vector space with inner product $(\cdot,\cdot)$, and $V_i\subset V$.
The operator $A : V \to V $ is symmetric positive definite with respect to the inner product $(\cdot,\cdot)$.
On these spaces there are two operators, $Q_i: V \to V_i$, $A_i: V_i \to V_i$, where \begin{align*} (A_i v,w) &= (A v, w)~~~\forall w\in V_i \\ (Q_i v,w) &= ( v, w)~~~\forall w\in V_i. \end{align*}
Is it possible to prove the inequality $$ (Q_iu,Q_iu)\le \lambda (A_i^{-1}Q_iu,u), $$ for any $u\in V$, where $\lambda$ is the largest eigenvalue of $A$?
I am not sure how to introduce $\lambda$ into the inequality. I tried $$ (Q_iu,Q_iu) = (A_i A_i^{-1} Q_iu,Q_iu)= (A A_i^{-1} Q_iu,Q_iu). $$ The Cauchy-Schwarz inequality doesn't seem to help, or at least I don't see how it is applicable
By the first assumption the operator $A_i$ is positive definite. By the second assumption we have $u-Q_iu\perp V_i$ for any $u\in V.$ Therefore $$\langle A_i^{-1}Q_iu,u\rangle =\langle A_i^{-1}Q_iu,Q_iu\rangle+\langle A_i^{-1}Q_iu,u-Q_iu\rangle\\ =\langle A_i^{-1}Q_iu,Q_iu\rangle \ge \mu_i \langle Q_iu,Q_iu\rangle $$ where $\mu_i$ is the least eigenvalue of $A_i^{-1}.$ Hence $\mu_i=\lambda_i^{-1},$ where $\lambda_i$ is the largest eigenvalue of $A_i.$ However the largest eigenvalue of $A_i$ is less than or equal to the largest eigenvalue $\lambda$ of $A.$ Indeed for $v\in V_i$ we have $$\langle A_iv,v\rangle =\langle Av,v\rangle\le \lambda\|v\|^2$$ Summarizing we have obtained $$\langle A_i^{-1}Q_iu,u\rangle\ge \lambda_i^{-1}\langle Q_iu,Q_iu\rangle \ge \lambda^{-1}\langle Q_iu,Q_iu\rangle$$