Inequality with $(x+y)(y+z)(z+w)(w+x)=1$

Question

Let $x,y,z,w>0$ and such that $$(x+y)(y+z)(z+w)(w+x)=1.$$ Show that $$\sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}\le 2.$$

I'm trying to use Holder's inequality $$(\sqrt[3]{xyz}+\sqrt[3]{yzw})^3 \le (x+y)(y+z)(z+w)$$ $$(\sqrt[3]{zwx}+\sqrt[3]{wxy})^3\le (z+w)(w+x)(x+y)$$ so $$\sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}\le \sqrt[3]{(x+y)(y+z)(z+w)}+\sqrt[3]{(z+w)(w+x)(x+y)}.$$

Answer 1

Since $$\prod_{cyc}(x+y)-\sum_{cyc}x\sum_{cyc}xyz=(xz-yw)^2\geq0,$$ by Holder we obtain:

$$2=2\left(\prod_{cyc}(x+y)\right)^{\frac{1}{4}}\geq2\left(\sum_{cyc}x\sum_{cyc}xyz\right)^{\frac{1}{4}}=$$ $$=\sqrt[4]{\left(4^2\sum_{cyc}xyz\right)\sum_{cyc}x}\geq\sqrt[4]{\left(\sum_{cyc}\sqrt[3]{xyz}\right)^3\sum_{cyc}x}.$$ Thus, it's enough to prove that $$\sum_{cyc}x\geq\sum_{cyc}\sqrt[3]{xyz},$$ which is true by AM-GM: $$\sum_{cyc}x=\frac{1}{3}\sum_{cyc}(x+y+z)\geq\frac{1}{3}\sum_{cyc}3\sqrt[3]{xyz}=\sum_{cyc}\sqrt[3]{xyz}.$$ Done!

I used Holder for two sequences: $$16\sum_{cyc}xyz=\left(\sum_{cyc}1\right)^2\sum_{cyc}xyz\geq\left(\sum_{cyc}\sqrt[3]{1^2xyz}\right)^3=\left(\sum_{cyc}\sqrt[3]{xyz}\right)^3$$

Answer 2

This does not seem the best, but using only Holder's inequality, we can get $$\begin{eqnarray} S^{12}&\le& (x+y+x+y)(y+y+x+x)(y+y+x+x)(z+y+z+y)\\&&(y+z+z+y)(y+z+z+y)(z+w+z+w)(z+z+w+w)\\&&(z+z+w+w)(x+w+x+w)(x+w+w+x)(x+w+w+x) = 2^{12}\left(\prod_{\text{cyc}}(x+y)\right)^3 =2^{12}, \end{eqnarray}$$for $S= \sqrt[3]{xyz}+\sqrt[3]{yzw}+\sqrt[3]{zwx}+\sqrt[3]{wxy}.$