<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] 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!
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