Prove that $(x+(y+z^{1/4})^{1/3})^{1/2}\geqslant(xyz)^{1/32}$ Please help with the inequality $$\large\sqrt[2]{x+\sqrt[3]{y+\sqrt[4]{z}}}\geqslant\sqrt[32]{xyz}.$$ I've tried with Cauchy's theorem. And don't know what to do later.
 A: Hint: One needs to be able to lower bound sums $u+v^\theta$ with $u$ and $v$ nonnegative and $\theta$ in $(0,1)$, by some power of $uv$. Thus, you might want to show first the optimal inequality stating that $$u+v^\theta\geqslant(1+\theta)\cdot\theta^{-1/(1+\theta)}\cdot(uv)^{\theta/(1+\theta)},$$ and to apply this twice to your setting, once for $\theta=1/4$ then for $\theta=1/15$. If I am not mistaken the lower bound one gets in this way is $c\cdot(xyz)^{1/32}$ with $c\approx1.477228.$
A: Both sides are positive so the inequality is equivalent with $$x+\sqrt[3]{y+\sqrt[4]{z}} \geq \sqrt[16]{xyz}$$
Note that by the AM-GM inequality $$y+\sqrt[4]{z}=\frac14\sqrt[4]{z}+\frac14\sqrt[4]{z}+\frac14\sqrt[4]{z}+\frac14\sqrt[4]{z}+y \geq 5 \sqrt[5]{\frac{1}{256}yz}$$
Then note that $\sqrt[3]{y+\sqrt[4]{z}} \geq \sqrt[3]{5 \sqrt[5]{\frac{1}{256}yz}} = \sqrt[15]{\frac{3125}{256}yz}$. 
Then note that by the AM-GM inequality
$$x+\sqrt[3]{y+\sqrt[4]{z}} \geq x+15 \cdot \frac1{15} \sqrt[15]{\frac{3125}{256}yz} \geq 16 \sqrt[16]{\frac{3125}{256\cdot 15^{15}}xyz} = \sqrt[16]{\frac{3125\cdot 16^{16}}{256\cdot 15^{15}}xyz} = \sqrt[16]{\frac{3125\cdot 16^{14}}{15^{15}}xyz} > 1.477228 \sqrt[16]{xyz} > \sqrt[16]{xyz}$$
