<s>[INST] 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?
 [/INST] 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.
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