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It is well-known that $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-a-b-c\le 0$, then we have $$\begin{array}{rl}& (a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})+(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\\ =& (a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})+3(a^2+b^2+c^2)-(a+b+c{)}^2\\ =& (a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-a-b-c)+3(a^2+b^2+c^2)\\ \le & 3(a^2+b^2+c^2).\end{array}$$ 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 $$(x^2+y^2+z^2)(xy+yz+zx)+\sum x^4\ge 4(x^2{y}^2+y^2{z}^2+z^2{x}^2)$$ which is equivalent to $$\sum x^4+xyz\sum x+\sum xy(x^2+y^2)\ge 4\sum x^2{y}^2.(1)$$ By Schur's inequality, we have $$\sum x^2(x-y)(x-z)\ge 0$$ hence $$\sum x^4+xyz\sum x\ge \sum xy(x^2+y^2).(2)$$ Note that by Cauchy's inequality, $$\sum xy(x^2+y^2)\ge \sum 2x^2{y}^2,(3)$$ then the result is followed from (1), (2) and (3). $\square$