Question 72, Given $a, b, c, d \in R, ab + bc + cd + da = 1$, then the minimum possible value of $a^{2} + 2b^{2} + 3c^{2} + 4d^{2}$ is $\qquad$ -
Question 72, Solution: According to the mean inequality and Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
a^{2}+2 b^{2}+3 c^{2}+4 d^{2} \\
\geq 2 \sqrt{\left(a^{2}+3 c^{2}\right)\left(2 b^{2}+4 d^{2}\right)} \\
=2 \sqrt{\frac{\left(a^{2}+3 c^{2}\right)\left(1+\frac{1}{3}\right)}{\frac{4}{3}} \cdot \frac{\left(2 b^{2}+4 d^{2}\right)\left(\frac{1}{2}+\frac{1}{4}\right)}{\frac{3}{4}}} \\
\geq 2 \sqrt{|a+c| \cdot|b+d|}=2
\end{array}
$$

When $a=\frac{\sqrt{3}}{2}, c=\frac{\sqrt{3}}{6}, b=\frac{\sqrt{3}}{3}, d=\frac{\sqrt{3}}{6}$, equality holds.