I am trying to show the following statement: Let $H$ be a Hilbert space and $A:H \rightarrow H$ be $K$-Lipschitz continious and $\beta$-strongly monotone and let $f \in H$. (.,.) denotes the scalar product of the Hilbert space $H$. Further let $V$ be a finite dimensional subspace of $H$. The variational equation \begin{equation} (\forall y \in V) (Ax,y)=(f,y) ~~\text{has a unique solution}~~ \overline{x_v}. \end{equation} Further, the following inequality holds \begin{equation} \lVert \overline{x}-\overline{x_v}\rVert \leq \frac{K}{\beta}min_{y\in V}\lVert\overline{x}-y \rVert. \end{equation} Here $\overline{x}$ denotes the unique solution of \begin{equation} (\forall y \in H) (Ax,y)=(f,y) . \end{equation}
I know that the existence of the solution $\overline{x_v}$ follows from Lax-Milgram Theorem.Thus, the first part of the statement is clear to me. However, I am struggling to prove the inequality. I used the continuity and the monotonicity to show that \begin{equation} \lVert \overline{x}-\overline{x_v}\rVert ^2 \leq \frac{1}{\beta}\lVert\overline{x}-\overline{x_v}\rVert \lVert A\overline{x}-A\overline{x_v}\rVert \leq \frac{K}{\beta}\lVert\overline{x}-\overline{x_v}\rVert ^2. \end{equation} Now I do not understand how I can show that this is also true if I replace $\overline{x_v}$ with $y \in V$. In addition, I want to deduce Ceas Lemma from the inequality. Could anyone please help me to show the inequality and to deduce Ceas Lemma? Thanks!