Mathematical Reflections 2012, Issue 5 - Problem J245

Problem:
Find all triples $(x,y,z)$ of positive real numbers satisfying simultaneously the inequalities $x+y+z-2xyz \leq 1$ and $$xy+yz+zx+\dfrac{1}{xyz} \leq 4.$$

Proposed by Titu Andreescu.

Solution:
Using HM-AM Inequality, from the first inequality we have
$$\dfrac{9xyz}{xy+yz+zx} \leq x+y+z \leq 1+2xyz,$$ which gives $\dfrac{9xyz}{2xyz+1} \leq xy+yz+zx$. From the second inequality, we have $$\dfrac{9xyz}{2xyz+1}+\dfrac{1}{xyz} \leq 4.$$ Putting $t=xyz>0$ and clearing the denominators, we obtain $$9t^2+2t+1 \leq 4t(2t+1),$$ i.e. $(t-1)^2 \leq 0$, which gives $t=1$. So, the first inequality is $x+y+z \leq 3xyz$, but by AM-GM Inequality $3xyz \leq x+y+z$, which implies $x+y+z=3xyz$. The equality holds if and only if $x=y=z=1$, so the only triple which satisfies the two inequalities is $(1,1,1)$.