8. Let $a>0, b>0$. Then the inequality that does not always hold is ( ).
(A) $\frac{2}{\frac{1}{a}+\frac{1}{b}} \geqslant \sqrt{a b}$
(B) $\frac{1}{a}+\frac{1}{b} \geqslant \frac{4}{a+b}$
(C) $\sqrt{|a-b|} \geqslant \sqrt{a}-\sqrt{b}$
(D) $a^{2}+b^{2}+1>a+b$
8. A.
$$
\text { From } \frac{1}{a}+\frac{1}{b} \geqslant 2 \sqrt{\frac{1}{a b}} \Rightarrow \frac{2}{\frac{1}{a}+\frac{1}{b}} \leqslant \sqrt{a b} \text {. }
$$

Therefore, option (A) does not always hold.