How prove $ax(a+x)+by(b+y)+cz(c+z)\ge 3(abc+xyz)$ let $a,b,c,x,y,z\ge 0$ and such
$$a+b+c=x+y+z$$
show that
$$ax(a+x)+by(b+y)+cz(c+z)\ge 3(abc+xyz)$$but it does not help for a proof of the starting inequality (at least I don't see, how it helps).
I tried also BW, but we get there something, which impossible to kill during a competition.
 A: It's enough to prove that $x^2a+y^2b+z^2c+xyz\geq4abc$.
Let $x^2a+y^2b+z^2c+xyz<4abc$, $x=kp$, $y=kq$ and $z=kr$, where $k>0$ and 
$p^2a+r^2b+q^2c+pqr=4abc$.
Hence, $p^2a+r^2b+q^2c+pqr=4abc>x^2a+y^2b+z^2c+xyz=k^2(p^2a+r^2b+q^2c+kpqr)$,
which says that $0<k<1$ and $a+b+c=x+y+z=k(p+q+r)<p+q+r$, 
which is contradiction because we'll prove now that $a+b+c\geq p+q+r$,
where $a$, $b$, $c$, $p$, $q$ and $r$ are positives such that $p^2a+r^2b+q^2c+pqr=4abc$.
Indeed, we'll rewrite the last condition in the following form.
$$\frac{p^2}{4bc}+\frac{q^2}{4ac}+\frac{r^2}{4ab}+\frac{pqr}{4abc}=1.$$ 
Let $\cos\alpha=\frac{p}{2\sqrt{bc}}$,  $\cos\beta=\frac{q}{2\sqrt{ac}}$ and $\cos\gamma=\frac{r}{2\sqrt{ab}}$.
Hence, $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1,$$
which says that $\alpha+\beta+\gamma=180^{\circ}$ and we need to prove that
$$a+b+c\geq2\sqrt{bc}\cos\alpha+2\sqrt{ac}\cos\beta+2\sqrt{ab}\cos\gamma.$$
Let $\Delta ABC$ is a triangle such that $\measuredangle A=\alpha$, $\measuredangle B=\beta$ and $\measuredangle C=\gamma$, 
$\vec{u}\uparrow\uparrow \vec{CB}$, $|\vec{u}|=\sqrt{a}$, $\vec{v}\uparrow\uparrow \vec{BA}$, $|\vec{v}|=\sqrt{b}$, $\vec{w}\uparrow\uparrow \vec{AC}$, $|\vec{w}|=\sqrt{c}$ and since
$$(\vec{u}+\vec{v}+\vec{w})^2\geq0,$$
we are done!
