<s>[INST] Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$ 
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$

It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?
 [/INST] 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
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