Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$

Yet another way in which this can be shown using Schur's inequality in tandem with the AM-GM inequality is as follows: $$ a^2+b^2+c^2+3(a^2b^2c^2)^{1/3}\geqslant a^{2/3}b^{4/3} + a^{4/3}b^{2/3} +b^{2/3}c^{4/3} + b^{4/3}c^{2/3} + a^{2/3}c^{4/3} + a^{4/3}c^{2/3} \\[2ex]= 2\left({a^{2/3}b^{4/3} + a^{4/3}b^{2/3}\over 2} + {b^{2/3}c^{4/3} + b^{4/3}c^{2/3}\over 2} + {a^{2/3}c^{4/3} + a^{4/3}c^{2/3}\over 2}\right)\\[2ex] \geqslant 2(ab + bc + ac) $$

Let $a=x^3$, $b=y^3$, $c=z^3$, then it can be rewritten as: $$ x^6+y^6+z^6+3 x^2 y^2 z^2-2 \left(x^3 y^3+x^3 z^3+y^3 z^3\right)\geq 0 $$ Use the following notations: $$S_{3}:=xyz\qquad S_2:=xy+yz+xz\qquad S_1=x+y+z$$ Then: $$ x^6+y^6+z^6=S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-12 S_2 S_3 S_1-2 S_2^3+3 S_3^2 $$ $$ x^3 y^3+x^3 z^3+y^3 z^3=S_2^3-3 S_1 S_3 S_2+3 S_3^2 $$ $$ 3x^2y^2z^2=3S_3^2 $$ Then we only have to prove: $$ S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-6 S_2 S_3 S_1-4 S_2^3\geq 0 $$ Now put $S_2=S_1^2$, and notice that with this: $$ \left.S_1^6-6 S_2 S_1^4+6 S_3 S_1^3+9 S_2^2 S_1^2-6 S_2 S_3 S_1-4 S_2^3\right|_{S_2=S_1^2}=0 $$ Thus this can be factorised as: $$ \left(S_1^2-S_2\right) \left(S_1^4-5 S_2 S_1^2+6 S_3 S_1+4 S_2^2\right)\geq0 $$ Since: $(x+y+z)^2\geq 3(xy+yz+xz)\Rightarrow S_1^2\geq 3S_2$ by rearrangement, it is enough to prove that the second factor is non-negative. Return to our previous notations, enough to show: $$ x^4+y^4+z^4+(x+y+z)xyz-x^3y-y^3x-y^3z-z^3y-x^3z-xz^3= $$ $$ =x^2(x-y)(x-z)+y^2(y-x)(y-z)+z^2(z-x)(z-y)\geq 0 $$ Which is trivially true by applying Schur's inequality

$x^3=a^2,y^3=b^2,z^3=c^2 \implies x^3+y^3+z^2 +3xyz \ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$

we have $x^3+y^3+z^3 +3xyz \ge xy(x+y)+yz(y+z)+xz(x+z)$

$xy(x+y)\ge 2xy\sqrt{xy}=2\sqrt{(xy)^3} \implies xy(x+y)+yz(y+z)+xz(x+z)\ge 2(\sqrt{(xy)^3}+\sqrt{(yz)^3}+\sqrt{(xz)^3})$