<s>[INST] How prove that $ x+y+z>4$ for $ a+b+c=4$ and $ ax+by+cz=xyz$? Given positive reals $ a,b,c,x,y,z$ such that $ a+b+c=4$ and $ ax+by+cz=xyz$.
How prove that $ x+y+z>4$?
 [/INST] Suppose there exist $a, b, c, x, y, z > 0$ such that
$$a+b+c=4,$$
$$ax+by+cz = xyz,$$
$$x+y+z \leq 4.$$
Observe that we have
$$axyz + bxyz + cxyz = 4xyz,$$
so that
$$axyz + bxyz + cxyz = 4ax + 4by + 4cz,$$
and hence
$$ax(yz-4) + by(xz-4) + cz(xy-4) = 0.$$
Since
$x+y+z \leq 4$
and
$a, b, c, x, y, z > 0,$
what follow are the inequalities
$$xy, yz, xz < 4,$$
whence
$$ax(yz-4) + by(xz - 4) + cz(xy-4) < 0;$$
a contradiction, qed.
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