We know that $a^{2}+b^{2}+c^{2} \geq a b+b c+c a$ and thus $3\left(a^{2}+b^{2}+c^{2}\right) \geq(a+b+c)^{2}$. Due to the inequality of the geometric and arithmetic mean, we have $a+b+c \geq 3(a b c)^{\frac{1}{3}}$ and on the other hand, it is given that $a+b+c \geq a b c$. Therefore, we have two inequalities:

$$\begin{aligned} & a^{2}+b^{2}+c^{2} \geq \frac{1}{3}(a+b+c)^{2} \geq 3(a b c)^{\frac{2}{3}} \\ & a^{2}+b^{2}+c^{2} \geq \frac{1}{3}(a+b+c)^{2} \geq \frac{1}{3}(a b c)^{2} \end{aligned}$$

We raise the first inequality to the power of $\frac{3}{4}$ and the second to the power of $\frac{1}{4}$ (and we are allowed to do this because everything is positive):

$$\begin{aligned} & \left(a^{2}+b^{2}+c^{2}\right)^{\frac{3}{4}} \geq 3^{\frac{3}{4}}(a b c)^{\frac{1}{2}} \\ & \left(a^{2}+b^{2}+c^{2}\right)^{\frac{1}{4}} \geq 3^{-\frac{1}{4}}(a b c)^{\frac{1}{2}} \end{aligned}$$

Now we multiply these two and then we get

$$a^{2}+b^{2}+c^{2} \geq 3^{\frac{1}{2}}(a b c)$$

which is what we wanted.