Find maximize of $P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$ Let $x,y,z\in \mathbb{R^+}$ such that $6x+3y+2z=xyz$. Find maximize of $$P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$$

We will prove $$P\le \sqrt{\frac{16}{27}}$$
$$\Leftrightarrow \frac{1}{\sqrt{1+\frac{1}{x^2}}\sqrt[4]{\left(1+\frac{4}{y^2}\right)\left(1+\frac{9}{z^2}\right)}}\le \sqrt{\frac{16}{27}}$$
$$\Leftrightarrow \frac{1}{\sqrt[4]{\frac{\left(2x+y\right)^3\left(3y+2z\right)^2\left(3x+z\right)^3}{x^6y^5z^5}}}\le \sqrt{\frac{16}{27}}$$
$$\Leftrightarrow \left(2x+y\right)^3\left(3y+2z\right)^2\left(3x+z\right)^3\left(xyz\right)^4\ge \frac{729}{256}x^6y^5z^5\left(6x+3y+2z\right)^4$$
I don't know how to solve it now. Help me.

Also, $uvw$ helps.
Indeed, in my previous solution it's enough to prove that:
$$64(2a+b)^3(2a+c)^3(b+c)^2\geq729(2a+b+c)^4a^2bc.$$
Now, let $b+c=2u$ and $bc=v^2$, where $v>0$.
Thus, by AM-GM  $u\geq v$ and we need to prove that $f(u)\geq0,$ where
$$f(u)=16u^2(4a^2+4ua+v^2)^3-729(a+u)^4a^2v^2.$$
But, $$f'(u)=32u(4a^2+4ua+v^2)^3+192u^2a(4a^2+4ua+v^2)^2-4\cdot729(a+u)^3a^2v^2=$$
$$=4(512ua^6+9(256u^2-81v^2)a^5+3u(1024u^2-601v^2)a^4+5u^2(256u^2-207v^2)a^3+$$
$$+3uv^2(13u^2+32v^2)a^2+144u^2v^4a+8uv^6)>0,$$
which says that $f$ increases.
Thus, it's enough to prove our inequality for a minimal value of $u$, which happens for $u=v$,
id est, for $b=c$.
Since our inequality is homogeneous, it's enough to assume that $b=c=1$ and we need to prove that:
$$16(4a^2+4a+1)^3\geq729(a+1)^4a^2$$ or
$$4(2a+1)^3\geq27(a+1)^2a,$$ which is true by AM-GM:
$$4(2a+1)^3=4\left(a+2\cdot\frac{a+1}{2}\right)^3\geq4\left(3\sqrt[3]{a\left(\frac{a+1}{2}\right)^2}\right)^3=27(a+1)^2a.$$
