If $\sqrt{xy}+\sqrt[3]{xyz}<a(x+4y+4z)$ and $x,y,z>0$ then $a$ is

with the help of am gm inequality

$\displaystyle \sqrt{xy}\leq \left(\frac{x+y}{2}\right)$ and $\displaystyle \sqrt[3]{xyz}\leq \left(\frac{x+y+z}{3}\right)$

so $\displaystyle \sqrt{xy}+\sqrt[3]{xyz}\leq \frac{x+y}{2}+\frac{x+y+z}{3} = \frac{5x+5y+2z}{6}$

want be able to go further, could some help me


We'll prove that $$x+4y+4z\geq3\left(\sqrt{xy}+\sqrt[3]{xyz}\right).$$ Indeed, by AM-GM $$4z+x+4y-3\sqrt{xy}=4z+2\cdot\frac{x+4y-3\sqrt{xy}}{2}\geq$$ $$\geq3\sqrt[3]{4z\left(\frac{x+4y-3\sqrt{xy}}{2}\right)^2}=3\sqrt[3]{z\left(x+4y-3\sqrt{xy}\right)^2}.$$ Thus, it remains to prove that $$\left(x+4y-3\sqrt{xy}\right)^2\geq xy,$$ which is $$(\sqrt{x}-2\sqrt{y})^2(x+4y-2\sqrt{xy})\geq0.$$ The equality occurs for $x=4y$ and $y=4z$.

Id est, $a>\frac{1}{3}$ are all values of $a$, for which the inequality $$\sqrt{xy}+\sqrt[3]{xyz}<a(x+4y+4z)$$ is true for all positives $x$, $y$ and $z$.


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