Proof: Note that, if $(a, b, c, d)$ is in increasing order, then
$$\frac{1}{b+c+d} \geq \frac{1}{c+d+a} \geq \frac{1}{d+a+b} \geq \frac{1}{a+b+c}$$

Therefore, by Chebyshev's inequality, we have
$$4 L H S \geq\left(\sum_{\text {cyc }} a^{2}\right)\left(\sum_{c y c} \frac{1}{b+c+d}\right) \geq \frac{16\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}{3(a+b+c+d)} \geq \frac{4 \sqrt{4\left(a^{2}+b^{2}+c^{2}+d^{2}\right)}}{3}$$

That is $\square$
$$\frac{a^{2}}{b+c+d}+\frac{b^{2}}{c+d+a}+\frac{c^{2}}{d+a+b}+\frac{d^{2}}{a+b+c} \geq \frac{4}{3}$$