<s>[INST] An inequality with condition I have a new inequality this is the following :
Let $x,y,z$ be real strictly positive number such as : 

$$-2 = - x y z +  x +  y +  z $$

Then we have :

$$\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{y}}+\sqrt{\frac{1}{z}}\geq \frac{9}{\sqrt{15+((x+1)(y+1)(z+1))^{(\frac{1}{3})}}}$$

My try :
I put the following substitution :
$a^3=x$;
$b^3=y$;
$c^3=z$
And I have tried Holder (a generalized version) to get :
$$((a^3+1)(b^3+1)(c^3+1))^{(\frac{1}{3})}\geq abc+1 $$
The inequality becomes :

$$\sqrt{\frac{1}{a^3}}+\sqrt{\frac{1}{b^3}}+\sqrt{\frac{1}{c^3}}\geq \frac{9}{\sqrt{15+(abc+1)}}$$
  And 
$$-2=a^3b^3c^3+a^3+b^3+c^3$$

After that I'm stuck...There is someone to achieve this ? 
 [/INST] The condition gives $\sum\limits_{cyc}\frac{1}{x+1}=1.$
Let $\frac{1}{x+1}=\frac{a}{3},$ $\frac{1}{y+1}=\frac{b}{3}$ and $\frac{1}{z+1}=\frac{c}{3}.$
Hence, $a+b+c=3$ and we need to prove that
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{15+\frac{3}{\sqrt[3]{abc}}}}$$ or
$$\sum_{cyc}\sqrt{a(a+b)(a+c)}\geq\sqrt{\frac{81\prod\limits_{cyc}(a+b)}{15+\frac{a+b+c}{\sqrt[3]{abc}}}}.$$
Now, by C-S
$$\sum_{cyc}\sqrt{a(a+b)(a+c)}=\sqrt{\sum_{cyc}\left(a(a+b)(a+c)+2\sqrt{a(a+b)(a+c)}\sqrt{b(a+b)(b+c)}\right)}=$$
$$=\sqrt{\sum_{cyc}\left(a^3+a^2b+a^2c+abc+2\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\left(a^3+a^2b+a^2c+abc+2(ab(a+b+c)+abc)\right)}=$$
$$=\sqrt{\sum_{cyc}\left(a^3+3a^2b+3a^2c+5abc\right)}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\left(a^3+3a^2b+3a^2c+5abc\right)\geq\frac{81\prod\limits_{cyc}(a+b)}{15+\frac{a+b+c}{\sqrt[3]{abc}}}.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$27u^3+9w^3\geq\frac{81(9uv^2-w^3)}{15+\frac{3u}{w}},$$
which is a linear inequality of $v^2$.
Hence, it's enough to prove the last inequality for an extreme value of $v^2$,
which happens for an equality case of two variables.
Since the last inequality is homogeneous, we can assume that $b=c=1$.
Also, let $a=t^3$.
Id est, we need to prove that
$$(t^3+15t+2)(t^9+6t^6+21t^3+8)\geq162t(t^3+1)^2$$ or
$$(t-1)^2\left(t^{10}+2t^9+18t^8+42t^7+66t^6+18t^5+3t^4-12t^3-36t^2-10t+16\right)\geq0,$$
which is very strong, but  true.
</s>