Prove $x+y+z \ge xy+yz+zx$ Given $x,y,z \ge 0$ and $x+y+z=4-xyz$ Then Prove that
$$x+y+z \ge xy+yz+zx$$
My try:
Letting $x=1-a$, $y=1-b$ and $z=1-c$ we get
$$(1-a)+(1-b)+(1-c)+(1-a)(1-b)(1-c)=4$$
$$-(a+b+c)-(a+b+c)+ab+bc+ca-abc=0$$
$$ab+bc+ca-abc=2(a+b+c)$$
Where $a, b,c \le 1$
is there a clue here?

Suppose otherwise $x+y+z<xy+xz+yz$. Let $x=ka$, $y=kb$ and $z=kc$, such that $k>0$ and $a+b+c=ab+ac+bc$.
Thus, $$k(a+b+c)<k^2(ab+ac+bc),$$ which gives $$k>1$$ and
$$4=k(a+b+c)+k^3abc>a+b+c+abc,$$ which is a contradiction because we'll prove now that
$$a+b+c+abc\geq4.$$
Indeed, we need to prove that
$$\frac{(ab+ac+bc)^2}{a+b+c}+abc\geq\frac{4(ab+ac+bc)^3}{(a+b+c)^3}$$ or
$$\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3+3a^4bc-a^3b^2c-a^3c^2b-a^2b^2c^2)\geq0,$$ which is true by Muirhead.
