If $x \in[0,1]$, the inequality $2 \log _{2}(2 x+b) \geqslant \log _{2}(x+1)$ always holds for $x$, find the range of real number $b$.

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The original problem in English:

If $x \in[0,1]$, the inequality $2 \log _{2}(2 x+b) \geqslant \log _{2}(x+1)$ always holds for $x$, find the range of real number $b$.
From the condition, we have $(2 x+b)^{2} \geqslant x+1$, so $4 x^{2}+(4 b-1) x+b^{2}-1 \geqslant 0$. Also, $b > -2 x, x \in [0,1]$, so $b > (-2 x)_{\max }=0$.

Method one: Taking $x=0$ gives $b^{2} \geqslant 1$, so $b \geqslant 1$ (necessary condition). When $b \geqslant 1$, $4 x^{2}+(4 b-1) x+b^{2}-1 \geqslant 0$ is obviously true, so $b \geqslant 1$ is a necessary and sufficient condition.
Method two: Let $f(x)=4 x^{2}+(4 b-1) x+b^{2}-1$, and on $[0,1]$, $f(x) \geqslant 0$ always holds, $b \geqslant \frac{17}{8}$ or $b \geqslant 1$. Therefore, $b \geqslant 1$.
Method three: From $4 x^{2}+(4 b-1) x+b^{2}-1 \geqslant 0$, we get $b^{2}+4 x b+4 x^{2}-x-1 \geqslant 0$. Since $b>0$, we have $b \geqslant -2 x+\sqrt{x+1}, x \in [0,1]$, which means $b \geqslant (-2 x+\sqrt{x+1})_{\text {max }}, x \in [0,1]$. Let $S=-2 x+\sqrt{x+1} . u=\sqrt{x+1}$, then $u \in [1, \sqrt{2}], S=-2 u^{2}+u+2=-2\left(u-\frac{1}{4}\right)^{2}+2+\frac{1}{8}$, so $S_{\text {max }}=1(u=1$ is achieved), so $b \geqslant 1$.