51 .
$$\begin{array}{l} \sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)} \geqslant \\ a b c+\sqrt[3]{\left(a^{3}+a b c\right)\left(b^{3}+a b c\right)\left(c^{3}+a b c\right)} \Leftrightarrow \\ \sqrt{\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right)} \geqslant \\ 1+\sqrt[3]{\left(\frac{a^{2}}{b c}+1\right)\left(\frac{b^{2}}{c a}+1\right)\left(\frac{c^{2}}{a b}+1\right)} \end{array}$$

Let $x=\frac{a}{c}, y=\frac{b}{a}, z=\frac{c}{b}$, then $x y z=1$,
$$(1) \Leftrightarrow \sqrt{(x+y+z)(x y+y z+z x)} \geqslant 1+\sqrt[3]{\left(\frac{x}{z}+1\right)\left(\frac{y}{x}+1\right)\left(\frac{z}{y}+1\right)}$$

Notice that $x y z=1$, we have
$$\left(\frac{x}{z}+1\right)\left(\frac{y}{x}+1\right)\left(\frac{z}{y}+1\right)=(x+y)(-y+z)(z+x)$$

Also,
$$\begin{aligned} (x+y+z)(x y+y z+z x)= & (x+y)(y+z)(z+x)+x y z= \\ & (x+y)(y+z)(z+x)+1 \end{aligned}$$

So we only need to prove
$$\sqrt{(x+y+z)(x y+y z+z x)} \geqslant 1+\sqrt[3]{(x+y)(y+z)(z+x)}$$

Proof 1: Let $p=\sqrt[3]{(x+y)(-y+z)(z+x)}$, so inequality (2) $\Leftrightarrow \sqrt{p^{3}+1} \geqslant 1+p \Leftrightarrow p(p+1)(p-2) \geqslant 0$, we only need to prove $p \geqslant 2$. By the AM-GM inequality, we get $(x+y)(y+z)(z+x) \geqslant 2 \sqrt{x y} \cdot 2 \sqrt{y z} \cdot 2 \sqrt{z x}=8 x y z=8$, so, $p \geqslant 2$.

Proof 2: First prove the strengthened inequality: $\sqrt{(x+y+z)(x y+y z+z x)} \geqslant 1+$ $\frac{1}{3}\left(\frac{y+z}{\sqrt{y z}}+\frac{z+x}{\sqrt{z x}}+\frac{x+y}{\sqrt{x y}}\right)$.
By the Cauchy-Schwarz inequality and noting that $x y z=1$, we get
$$[x+(y+z)][y z+x(y+z)] \geqslant[\sqrt{x y z}+\sqrt{x}(y+z)]^{2}=\left(1+\frac{y+z}{\sqrt{y z}}\right)^{2}$$

That is,
$$\sqrt{(x+y+z)(x y+y z+z x)} \geqslant 1+\frac{y+z}{\sqrt{y z}}$$

Similarly,
$$\begin{array}{l} \sqrt{(x+y+z)(x y+y z+z x)} \geqslant 1+\frac{z+x}{\sqrt{z x}} \\ \sqrt{(x+y+z)(x y+y z+z x)} \geqslant 1+\frac{x+y}{\sqrt{x y}} \end{array}$$

So, adding them up, we get
$$\sqrt{(x+y+z)(x y+y z+z x)} \geqslant 1+\frac{1}{3}\left(\frac{y+z}{\sqrt{y z}}+\frac{z+x}{\sqrt{z x}}+\frac{x+y}{\sqrt{x y}}\right)=$$

Then by the AM-GM inequality, we get
$$\frac{1}{3}\left(\frac{y+z}{\sqrt{y z}}+\frac{z+x}{\sqrt{z x}}+\frac{x+y}{\sqrt{x y}}\right) \geqslant \sqrt[3]{\frac{(x+y)(y+z)(z+x)}{\sqrt{x y} \cdot \sqrt{y z} \cdot \sqrt{z x}}}=$$