Find all positive real numbers such that
$$x^2+y^2+z^2=xyz+4$$
And $$xy+yz+zx=2(x+y+z)$$.
I substitute $x,y,z$ by $(a+\frac{1}{a}),(b+\frac{1}{b}),(c+\frac{1}{c})$ respectively where $abc=1$ which comes from the first relation. Putting the values in the second relation it becomes complex.
I tried another way. Multiplying second Equation by 2 and adding with the first equation we get ,$(x+y+z-2)^2=xyz+8$.I failed to find any method. Any ideas guys?
Best Answer
After homogenization we obtain: $$\frac{(x^2+y^2+z^2)(xy+xz+yz)}{2(x+y+z)}=xyz+\frac{(xy+xz+yz)^3}{2(x+y+z)^3}.$$ We'll prove that $$\frac{(x^2+y^2+z^2)(xy+xz+yz)}{2(x+y+z)}\geq xyz+\frac{(xy+xz+yz)^3}{2(x+y+z)^3}.$$ Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$, and $xyz=w^3$.
We see that it's enough to prove this inequality for a maximal value of $w^3$, which happens for equality case of two variables.
Since the inequality is symmetric and homogeneous, we can assume $y=z=1$, which gives $$(2x+7)(x^2-1)^2\geq0.$$ The equality occurs for $x=y=z,$ which gives $$3x^2=6x$$ or $$x=y=z=2.$$
Another way.
We need to prove that: $$\frac{(9u^2-6v^2)\cdot3v^2}{6u}\geq w^3+\frac{27v^6}{54u^3}.$$ Now, by AM-GM $$v^4\geq uw^3,$$ which says that it's enough to prove that $$\frac{(9u^2-6v^2)v^2}{2u}\geq\frac{v^4}{u}+\frac{v^6}{2u^3}$$ or $$9u^4-8u^2v^2-v^4\geq0$$ or $$(u^2-v^2)(9u^2+v^2)\geq0$$ and since $$u^2-v^2\geq0$$ it's $$\sum_{cyc}(x-y)^2\geq0,$$ our inequality is true.