Example $\mathbf{3}$ If the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies for all real numbers $x$,
$$
\sqrt{2 f(x)}-\sqrt{2 f(x)-f(2 x)} \geqslant 2
$$

then the function $f$ is said to have property $P$. Find the largest real number $\lambda$, such that if the function $f$ has property $P$, then for all real numbers $x$, $f(x) \geqslant \lambda$ always holds.
Solve: From equation (1) we get
$$
\begin{array}{l}
f(x) \geqslant 2, \\
\sqrt{2 f(x)}-2 \geqslant \sqrt{2 f(x)-f(2 x)} .
\end{array}
$$

Squaring both sides of equation (2) we get
$$
\begin{array}{l}
f(2 x) \geqslant 4 \sqrt{2 f(x)}-4 \\
\Rightarrow f(x) \geqslant 4, \text { and } f(x) \geqslant 4 \sqrt{2 f\left(\frac{x}{2}\right)}-4 .
\end{array}
$$

Conjecture:
$f(x)=c$ has property $P \Leftrightarrow \sqrt{2 c}-\sqrt{c} \geqslant 2$
$$
\Leftrightarrow c \geqslant 12+8 \sqrt{2} \text {. }
$$

It is easy to verify that $f(x)=12+8 \sqrt{2}$ has property $P$, thus $12+8 \sqrt{2} \geqslant \lambda$.
Next, we prove: If function $f(x)$ has property $P$, then for all real numbers $x$,
$$
f(x) \geqslant 12+8 \sqrt{2} \text {. }
$$

Define the sequence $\left\{a_{n}\right\}: a_{0}=2, a_{n+1}=4 \sqrt{2 a_{n}}-4$. We can prove: (1) $a_{n}<a_{n+1}$.
In fact,
$$
\begin{array}{l}
a_{n+1}-a_{n}=4 \sqrt{2 a_{n}}-4 \sqrt{2 a_{n-1}} \\
=4 \sqrt{2}\left(\sqrt{a_{n}}-\sqrt{a_{n-1}}\right) .
\end{array}
$$

By mathematical induction, it is easy to get $a_{n}<a_{n+1}$.
(2) $a_{n}<12+8 \sqrt{2}$.

When $n=0$, $a_{0}=2<12+8 \sqrt{2}$.
Assume when $n=k$, $a_{k}<12+8 \sqrt{2}$.
Then $a_{k+1}=4 \sqrt{2 a_{k}}-4$
$$
<4 \sqrt{24+16 \sqrt{2}}-4=12+8 \sqrt{2} \text {. }
$$

In summary, the sequence $\left\{a_{n}\right\}$ is monotonically increasing and bounded above. Therefore, it has a limit.
$$
\text { Let } \lim _{n \rightarrow+\infty} a_{n}=a \text {. }
$$

Then $a=4 \sqrt{2 a}-4 \Rightarrow a=12+8 \sqrt{2}$.
(3) For any non-negative integer $n$ and any real number $x$, $f(x) \geqslant a_{n}$.
When $n=0$, $f(x) \geqslant 2=a_{0}$.
Assume the proposition holds for $n$, i.e., for any real number $x$, $f(x) \geqslant a_{n}$.
Thus $f(x) \geqslant \sqrt[4]{2 f\left(\frac{x}{2}\right)}-4$
$\geqslant 4 \sqrt{2 a_{n}}-4=a_{n+1}$.
Let $n \rightarrow+\infty$.
Then $f(x) \geqslant \lim _{n \rightarrow+\infty} a_{n+1}=a=12+8 \sqrt{2}$.
In summary, $\lambda_{\max }=12+8 \sqrt{2}$.