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?
 A: 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.
A: Suppose $x+y+z < xy+yz+zx$, then from Schur's inequality we have
$$\begin{align}
\frac{9xyz}{x+y+z} &\geqslant 4(xy+yz+zx)- (x+y+z)^2 \\
&> (x+y+z)\left( 4- (x+y+z)\right) \\
&= (x+y+z) \cdot xyz \\
\end{align}$$
This gives $x+y+z< 3$, further we have from $4= x+y+z+xyz \geqslant 4\sqrt{xyz} \implies xyz \leqslant 1$, so $4 = x+y+z+xyz < 3+1$, a contradiction.
Equality is when $x=y=z=1$ or when $(x, y, z)=(2, 2, 0)$ or a permutation.
A: Also, we can use $uvw$ here.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition does not depend on $v^2$ and it's enough to prove our inequality for a maximal value of $v^2$, which happens for equality case of two variables.
Let $y=x$.
Thus, the condition gives $z=\frac{4-2x}{1+x^2},$ where $0\leq x\leq2$ and we need to prove that
$$2x+\frac{4-2x}{1+x^2}\geq x^2+2x\cdot\frac{4-2x}{1+x^2}$$ or
$$(2+x)(2-x)(x-1)^2\geq0$$ and we are done!
