Problem 1374

6. The inequality to be proved is equivalent to
$$2(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \geqslant 3\left(a^{3}+b^{3}+c^{3}+a b c\right)$$

Let $2 x=-a+b+c, 2 y=a-b+c, 2 z=a+b-c$, then $a=y+z, b=z+x, c=x+y, (1) \Leftrightarrow$
Thus, $a+b+c=2(x+y+z), a^{2}+b^{2}+c^{2}=2\left(x^{2}+y^{2}+z^{2}+x y+y z+z x\right)$, $a^{3}+b^{3}+c^{3}=2\left(x^{3}+y^{3}+z^{3}\right)+3\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\right), a b c=$ $x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}+2 x y$
The left side of (1) $=8(x+y+z)\left(x^{2}+y^{2}+z^{2}+x y+y z+z x\right)=8\left(x^{3}+y^{3}+z^{3}\right)+$ $16\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\right)+24 x y z$
The right side of (1) $=6\left(x^{3}+y^{3}+z^{3}\right)+12\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\right)+6 x y z$, so the inequality clearly holds.