A difficulty in obtaining a formula in the proof of a theorem. The theorem and part of its proof is given below:




But the last line in the proof did not come up with me after calculation, I do not know where I will use that $\lambda = <Ax_{0},x_{0}>$ , could anyone explain it for me please?
Also in the rest of the proof below:



I can not see how this $\alpha$ give us the final line, could anyone explain this for me please?  
 A: $$\langle A(x_0+\alpha v), x_0+\alpha v \rangle \geq  \lambda \langle x_0+\alpha v, x_0+\alpha v \rangle\\
\langle Ax_0, x_0 \rangle+\langle Ax_0, \alpha v \rangle+\langle A\alpha v, x_0 \rangle+\langle A\alpha v, \alpha v \rangle \geq  \lambda \left( \langle x_0, x_0 \rangle +\langle x_0, \alpha v \rangle+\langle \alpha v, x_0 \rangle+\langle \alpha v, \alpha v \rangle\right)\\
\lambda +\bar{\alpha}\langle Ax_0,  v \rangle+\alpha \langle A v, x_0 \rangle+|\alpha|^2\langle Av ,  v \rangle \geq  \lambda \left( 1 +\bar{\alpha}\langle x_0,  v \rangle+\alpha\langle  v, x_0 \rangle+|\alpha|^2\langle v,  v \rangle\right)\\
\bar{\alpha}\langle Ax_0,  v \rangle+\alpha \langle A v, x_0 \rangle+|\alpha|^2\langle Av ,  v \rangle \geq  \lambda \left( \bar{\alpha}\langle x_0,  v \rangle+\alpha\langle  v, x_0 \rangle+|\alpha|^2\langle v,  v \rangle\right)\\
$$
Now, using the fact that $A$ is self adjoint and that $\lambda=\bar{\lambda}$, we get
$$\bar{\alpha}\langle Ax_0,  v \rangle+\alpha \langle  v, Ax_0 \rangle+|\alpha|^2\langle Av ,  v \rangle \geq   \bar{\alpha}\langle \lambda x_0,  v \rangle+\alpha\langle  v, \lambda x_0 \rangle+|\alpha|^2\langle \lambda v,  v \rangle\\
\overline{\alpha\langle  v, Ax_0 \rangle}+\alpha \langle  v, Ax_0 \rangle+|\alpha|^2\langle Av ,  v \rangle \geq   \overline{\alpha\langle  v, \lambda x_0 \rangle}+\alpha\langle  v, \lambda x_0 \rangle+|\alpha|^2\langle \lambda v,  v \rangle\\
$$
Now move everything on the left hand side and you obtain exactly the given inequality.
For the edit Taking $\alpha$ of the given form, the part inside the real part is real and your inequality becomes of the form
$$\gamma r+ \delta r^2 \geq 0$$
Now, unless $\gamma=0$, the LHS has two real roots and hence takes both positive and negative values.
A: Here is another approach (essentially the proof showing the
existence of a Lagrange multiplier):
Suppose $x_0$ is as above and $w_0$ is a unit vector that satisfies $x_0 \bot w_0$.
Note that $\|t x_0 + s w_0\|^2 = t^2+s^2$ and if we let $x(s) = \sqrt{1-s^2} x_0 + s w_0$ then $\|x(s) \| = 1$ for all $|s| <1$. In addition, $x(0) = x_0$ and $x'(0) = w_0$.
Now let $f(x) = \langle Ax, x \rangle$ and $\phi(s) = f(x(s))$.
Note that $\phi'(0) = 2 \operatorname{re} \langle Ax_0, w_0 \rangle$,
from which we conclude (since $x_0$ is a minimiser) that
$\langle Ax_0, w_0 \rangle = 0$ for all $w_0 \bot x_0$. In 
particular, $A x_0 \in \operatorname{sp} \{ x_0 \}$, hence $x_0$
is an eigenvector.
