Let $x,y,z \in \mathbb{R}$ and $x^2+y^2+z^2=3$. Prove that $$ (2-x^2+x)(2-y^2+y)(2-z^2+z) \leqslant 4 +x+y+z+xyz.$$
Observations
- At first, I think this inequality is easy. I applied AM-GM to the term $(2-x^2+x) \text{ }$,$(2-y^2+y) \text{ }$,$(2-z^2+z) \text{ }$ then try to rewrite the new inequality in term of $u=x+y+z \text{ }$,$w=xyz \text{ }$,$\text{ } v=xy+yz+zx$. Then I realized that the term $2-x^2+x$ can be negative. So the AM-GM approach fail.
- I then try to homogenize the inequality, and hope that after expand every thing. I can obtain some obvious inequality. Well, I fail to homogenize it.
- I try to assume $x\leqslant y \leqslant z$ and come up with some estimations but I get stuck.
hint:
$3(x^2+y^2+z^2) \ge (x+y+z)^2 \implies 3\ge x+y+z \ge -3,|xyz|^3 \le \sqrt{\dfrac{x^2+y^2+z^2}{3}}=1 \implies -1 \le xyz \le 1 \implies RHS \ge 0$
In Case LHS $<0$, the inequality is trivial true.
so you only need to consider LHS$ \ge 0$
there is two cases:
case 1: $2-x^2+x,2-y^2+y,2-z^2+z$ all positive. so you can use what you want.(but it is not easy also)
case2: two of them negtive, suppose $x<-1,y<-1$, then let $a=-x>1,b=-y>1,2<a+b \le \sqrt{2(3-z^2)} ,ab>1,|z|<1$
then RHS$ > f(z)>$LHS,try to find $f(z)$
another approach is expand LHS and replace with $uvw$ at both sides, then you will have to prove $f(w) \ge 0 \implies \Delta \le 0$, with $u^2-2v=3,|v|\le 3,|u|\le 3$, you should be able to get $\Delta \le 0$