Prove a geometric inequality, if 3 numbers are satisfying the condition I just got this problem, but I have no idea on how to prove that.
Prove that if $x,y,z\in\mathbb{R},\ x,y,z\ge 0$ and $2\cdot(x\cdot z+x\cdot y+y\cdot z)+3\cdot x\cdot y\cdot z = 9$, then $(\sqrt x + \sqrt y + \sqrt z )^4 \ge 72$. This is a geometric inequality.
Can anyone help me, please? Any kind of help (solutions, hints etc) is really appreciated. Thank you!
NOTE: I REALLY DON'T KNOW WHAT TITLE SHOULD I WRITE FOR THIS POST, SO PLEASE LEAVE A COMMENT IF YOU HAVE AN IDEA FOR THE POST TITLE. THANK YOU.

We need to prove that $x+y+z\geq\sqrt[4]{72}$, where $x$, $y$ and $z$ are non-negatives such that
$$\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=9$$
Let $x+y+z<\sqrt[4]{72}$, $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and $a+b+c=\sqrt[4]{72}$.
Hence, $k<1$ and $9=\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=k^4\sum\limits_{cyc}(2a^2b^2+k^2a^2b^2c^2)<$
$<\sum\limits_{cyc}(2a^2b^2+a^2b^2c^2)$, which is a contradiction because we'll prove now that 
$$\sum\limits_{cyc}(2a^2b^2+a^2b^2c^2)\leq9$$
Indeed, we need to prove that
$$9\left(\frac{a+b+c}{\sqrt[4]{72}}\right)^6\geq2(a^2b^2+a^2c^2+b^2c^2)\left(\frac{a+b+c}{\sqrt[4]{72}}\right)^2+3a^2b^2c^2$$ or
$$(a+b+c)^6\geq16(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2+144\sqrt2a^2b^2c^2$$
Since by AM-GM 
$(a+b+c)^4=(a^2+b^2+c^2+2(ab+ac+bc))^2\geq$
$\geq\left(2\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}\right)^2=8(a^2+b^2+c^2)(ab+ac+bc)$,
it remains to prove that
$$(a+b+c)^2(a^2+b^2+c^2)(ab+ac+bc)\geq2(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2+18\sqrt2a^2b^2c^2$$ or
$$(a+b+c)^2((a^2+b^2+c^2)(ab+ac+bc)-2(a^2b^2+a^2c^2+b^2c^2))\geq18\sqrt2a^2b^2c^2$$ or
$$(a+b+c)^2\sum\limits_{cyc}(a^3b+a^3c-2a^2b^2+a^2bc)\geq18\sqrt2a^2b^2c^2$$  or
$$(a+b+c)^2\sum\limits_{cyc}(ab(a-b)^2+a^2bc)\geq18\sqrt2a^2b^2c^2$$ 
Thus, it remains to prove that $(a+b+c)^3\geq18\sqrt2abc$, which is AM-GM:
$(a+b+c)^3\geq27abc\geq18\sqrt2abc$. Done!
