29. 16PQ $\leqslant a^{2}\left(-u^{2}+v^{2}+w^{2}\right)+b^{2}\left(u^{2}-v^{2}+w^{2}\right)+c^{2}\left(u^{2}+v^{2}-w^{2}\right) \Leftrightarrow$
$$\begin{array}{l} 16 P Q+2\left(a^{2} u^{2}+b^{2} v^{2}+c^{2} w^{2}\right) \leqslant \\ \left(a^{2}+b^{2}+c^{2}\right)\left(u^{2}+v^{2}+w^{2}\right) \end{array}$$

By Heron's formula $16 P^{2}=\left(a^{2}+b^{2}+c^{2}\right)^{2}-2\left(a^{4}+b^{4}+c^{4}\right)$, we get
$$16 P^{2}+2\left(a^{4}+b^{4}+c^{4}\right)=\left(a^{2}+b^{2}+c^{2}\right)^{2}$$

Similarly,
$$16 Q^{2}+2\left(u^{4}+v^{4}+w^{4}\right)=\left(u^{2}+v^{2}+w^{2}\right)^{2}$$

By equations (2), (3), and the Cauchy-Schwarz inequality, we get
$$\begin{array}{l} 16 P Q+2\left(a^{2} u^{2}+b^{2} v^{2}+c^{2} w^{2}\right) \leqslant \\ \sqrt{16 P^{2}+2\left(a^{4}+b^{4}+c^{4}\right)} \sqrt{16 Q^{2}+2\left(u^{4}+v^{4}+w^{4}\right)}= \\ \left(a^{2}+b^{2}+c^{2}\right)\left(u^{2}+v^{2}+w^{2}\right) \end{array}$$