Maximum value 1

Since the inequality holds true with $a=b=c=1$ $\Longrightarrow k\le \frac { 2 }{ 3 }$ .

Next, we'll show that $k=\frac { 2 }{ 3 }$ is achievable, hence $\frac { 2 }{ 3 }$ is the maximum value of $k$ .

When $k=\frac { 2 }{ 3 }$ , the inequality becomes:

${ a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 }+abc\left( a+b+c \right) \ge \frac { 2 }{ 3 } { \left( ab+bc+ca \right) }^{ 2 }\quad (1)$

${ \Longleftrightarrow 3\left( { a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 } \right) }\ge 2{ \left( { a }^{ 2 }{ b }^{ 2 }+{ b }^{ 2 }{ c }^{ 2 }+{ c }^{ 2 }{ a }^{ 2 } \right) +abc\left( a+b+c \right) }$

By applying AM-GM inequality,

$\left( { a }^{ 4 }+{ b }^{ 4 } \right) +\left( { b }^{ 4 }+{ c }^{ 4 } \right) +\left( { c }^{ 4 }+{ a }^{ 4 } \right) \ge 2{ a }^{ 2 }{ b }^{ 2 }+2{ b }^{ 2 }{ c }^{ 2 }+2{ c }^{ 2 }{ a }^{ 2 }\\ \Longrightarrow 3\left( { a }^{ 4 }+{ b }^{ 4 }+{ c }^{ 4 } \right) \ge 3\left( { a }^{ 2 }{ b }^{ 2 }+{ b }^{ 2 }{ c }^{ 2 }+{ c }^{ 2 }{ a }^{ 2 } \right) \quad (2)$

On the other hand,

${ a }^{ 2 }{ b }^{ 2 }+{ b }^{ 2 }{ c }^{ 2 }+{ c }^{ 2 }{ a }^{ 2 }-abc\left( a+b+c \right) =\frac { 1 }{ 2 } { \left( ab-bc \right) }^{ 2 }+\frac { 1 }{ 2 } { \left( bc-ca \right) }^{ 2 }+\frac { 1 }{ 2 } { \left( ca-ab \right) }^{ 2 }\ge 0\quad (3)$

From $(2)$ and $(3)$ , $(1)$ is proved.

So ${ k }_{ max }=\boxed { \frac { 2 }{ 3 } } .$