Unknown Inequality $$ \left( \sqrt{3}\sqrt{y(x+y+z)}+\sqrt{xz}\right)\left( \sqrt{3}\sqrt{x(x+y+z)}+\sqrt{yz}\right)\left( \sqrt{3}\sqrt{z(x+y+z)}+\sqrt{xy}\right) \leq 8(y+x)(x+z)(y+z)$$
I can prove this inequality, but i need know if this inequaliy is known...
 A: Here is a nice and simple proof (for $x,y,z\ge 0$ of course):
By Cauchy-Schwarz inequality:
$$\left( \sqrt{3}\sqrt{y(x+y+z)}+\sqrt{xz}\right)^2 \le (3+1)\big[(y(x+y+z) + xz\big] = 4(x+y)(y+z).$$
Similarly:
$$\left( \sqrt{3}\sqrt{x(x+y+z)}+\sqrt{yz}\right)^2 \le 4(x+z)(x+y)$$
$$\left( \sqrt{3}\sqrt{z(x+y+z)}+\sqrt{xy}\right)^2 \le 4(x+z)(y+z).$$
Taking the product of the above three inequalities we get the desired inequality.
A: We need to prove that
$$\prod\limits_{cyc}(a\sqrt{3(a^2+b^2+c^2)}+bc)\leq8\prod\limits_{cyc}(a^2+b^2)$$
which is true even for all reals $a$, $b$ and $c$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, it's obvious that the last inequality is equivalent yo $f(w^3)\leq0$, where $f$ is a convex function.
Hence, $f$ gets a maximal value for an extremal value of $w^3$.
We know that an equation $(x-a)(x-b)(x-c)=0$ or $x^3-3ux^2+3v^2x=w^3$ has three real roots $a$, $b$ and $c$.
Thus, a graph $y=w^3$ cross a graph $y=x^3-3ux^2+3v^2x$ in three points (maybe two of them coincide)
Let $u$ and $v^2$ be constants. 
Hence, $w^3$ gets an extremal value for an equality case of two variables.
Id est, it remains to prove our inequality for $b=c=1$, which gives 
$(x-1)^2(13x^2+34x+25)\geq0$, which is obvious. Done!
