It is well-known that $\sqrt{ab} + \sqrt{bc} + \sqrt{ca} - a - b - c \le 0$, then we have
$$\begin{aligned} & (a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) + (a-b)^2 + (b-c)^2 + (c-a)^2 \\ = & (a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) + 3(a^2+b^2+c^2) - (a+b+c)^2 \\ = & (a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-a-b-c) + 3(a^2+b^2+c^2) \\ \le & 3(a^2+b^2+c^2). \end{aligned}$$
To prove the other inequality, let $a = x^2$, $b = y^2$, $c = z^2$ where $x, y, z > 0$. The inequality can be rewritten as
$$(x^2 + y^2 + z^2)(xy + yz + zx) + \sum x^4 \ge 4(x^2y^2 + y^2z^2 + z^2x^2)$$
which is equivalent to
$$\sum x^4 + xyz \sum x + \sum xy(x^2 + y^2) \ge 4 \sum x^2 y^2. \quad (1)$$
By Schur's inequality, we have
$$\sum x^2(x-y)(x-z) \geq 0$$
hence
$$\sum x^4 + xyz \sum x \geq \sum xy(x^2 + y^2). \quad (2)$$
Note that by Cauchy's inequality,
$$\sum xy(x^2 + y^2) \geq \sum 2x^2y^2, \quad (3)$$
then the result is followed from (1), (2) and (3). $\Box$