Difficulty with: If $P,Q in mathcal{L}(H)$ and $0 leq P leq Q implies ||P|| leq ||Q||$

Question

In the book "Observation and Control for Operator Semigroups" of Marius Tucsnak and George Weiss page 392. I found the following statement
If $P,Q in mathcal{L}(H)$ and $0 leq P leq Q implies ||P|| leq ||Q|| quad (*) $
Which they said that we can get it easily using this one:  If $P in mathcal{L}(H)$ and $P geqslant 0$ then  $$|langle P x, yrangle|^{2} leqslantlangle P x, xrangle cdotlangle P y, yrangle quad forall x, y in H   (**)$$
as you can see in the attached picture: 
While thinking about this I thought that $(*)$ is obvious and we don't need any other statement (!) which seems to be not the case here.
I would like to ask for any hint or insight to get $(*)$ from $(**)$, and I would like also to see some example on where $(*)$ fail to be satisfied.
Thanks in advance!

Disintegrating By Parts · Accepted Answer

Let $y=Px$ in the inequality $|langle Px,yrangle|^2 le langle Px,xranglelangle Py,yrangle$:
$$
        |Px|^4 le langle Px,xranglelangle PPx,Pxrangle le langle Px,xrangle|P^2x||Px| le langle Px,xrangle|P||Px|^2 \
              |Px|^2 le |P|langle Px,xrangle \
            sup_{|x|=1}|Px|^2 le |P|sup_{|x|=1}langle Px,xrangle \
                 |P|^2le |P|sup_{|x|=1}langle Px,xrangle \
                 |P| le sup_{|x|=1}langle Px,xrangle
$$
The opposite inequality follows from $langle Px,xrangle le |Px||x|$. Therefore,
$$
                         |P|=sup_{|x|=1}langle Px,xrangle.
$$
That gives you what you want, starting with $0 le Ple Q$. For $|x|=1$,
$$
                      0 le langle Px,xrangle le langle Qx,xrangle le |Q| \
           |P| = sup_{|x|=1}langle Px,xrangle le |Q|.
$$

Difficulty with: If $P,Q in mathcal{L}(H)$ and $0 leq P leq Q implies ||P|| leq ||Q||$

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