with this inequality condition $xyz(x+y+z)=3$ Let $x,y,z$ be postive real numbers such that $xyz(x+y+z)=3$,show that
$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{54}{(x+y+z)^2}\ge 9$$
I found this problem on math magazine and the solution does not hold,see:

and I want to use Holder's inequality
$$\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{27}{(x+y+z)^2}$$
so we must show that the following inequality is true
$$\dfrac{81}{(x+y+z)^2}\ge 9$$
In fact, this also does not hold. Because $$xyz(x+y+z)=3\Longrightarrow x+y+z\ge 3$$
Now I suspect if $(1)$ is correct? Although I have not found any counter-example so far.

We'll prove a stronger inequality.
Let $a$, $b$ and $c$ be positives such that $ abc(a+b+c)=3$. Prove that:
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{72}{(a+b+c)^2}\ge 11$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $v^2$.
In another hand, $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{72}{(a+b+c)^2}=\frac{9v^4-6uw^3}{w^6}+\frac{8}{u^2}$ is an increasing function of $v^2$.
Thus, by $uvw$ (http://www.artofproblemsolving.com/community/c6h278791) it remains to check one case only: $b=a$ and $c=\sqrt{a^2+\frac{3}{a^2}}-a$, which gives
$\frac{2}{a^2}+\frac{1}{\left(\sqrt{a^2+\frac{3}{a^2}}-a\right)^2}+\frac{72}{\left(\sqrt{a^2+\frac{3}{a^2}}+a\right)^2}\geq11$ or
$146a^8+219a^4-99a^2+18\geq142a^6\sqrt{a^4+3}$.
Let $a^2=x$. Hence, we need to prove that
$(146x^4+219x^2-99x+18)^2\geq142^2x^6(x^2+3)$ or 
$(x-1)^2(128x^6+256x^5+768x^4-1932x^3+1281x^2-324x+36)\geq0$, which is obvious. 
