Functionnal analysis: Why $\langle AAx,x\rangle\underset{(*)}{\leq} (\|A\|+m)\langle Ax,x\rangle-\|A\|m ?$

Question

Let $(X,\langle\cdot ,\cdot \rangle)$ an inner vector space and $A\in \mathcal L(X)$ symetric such that $A\geq 0$.

I set $m=\inf\{\langle Ax,x\rangle \mid x\in X, \|x\|=1\}$ and thus $A-mI\geq 0$. If $\|x\|=1$, we have that

$$\mathbb R\ni\langle (A-mI)Ax,x\rangle\leq \|A\|\langle (A-mI)x,x\rangle=\|A\|\langle Ax,x\rangle-\|A\|m$$ and $$\|Ax\|^2=\langle AAx,x\rangle\underset{(*)}{\leq} (\|A\|+m)\langle Ax,x\rangle-\|A\|m.$$

I don't understand were $(*)$ come from. Notice: $\|A\|=\sup\{\|Ax\|\mid x\in X, \|x\|\leq 1\}$.

Answer 1

Note that

$$\langle (A-mI)Ax,x\rangle=\langle AAx,x\rangle-m\langle Ax,x\rangle,$$

so

$$\langle AAx,x\rangle-m\langle Ax,x\rangle \leq \|A\|\langle Ax,x\rangle-\|A\|m.$$

Since $\langle AAx,x\rangle = \|Ax\|^2$, because $A$ is symmetric, we have

$$ \|Ax\|^2-m\langle Ax,x\rangle \leq \|A\|\langle Ax,x\rangle-\|A\|m$$

then

$$ \|Ax\|^2 \leq (\|A\|+m)\langle Ax,x\rangle-\|A\|m.$$