$(!)\>(n-1)\sum_{k=1}^n\sqrt[3]{\frac{k}{n+1}}\leq n\sum_{k=1}^{n-1}\sqrt[3]{\frac{k}{n}}\Longleftrightarrow \sum_{k=1}^n\sqrt[3]{{k}}\leq \frac{n}{n-1}\sqrt[3]{\frac{n+1}{n}}\sum_{k=1}^{n-1}\sqrt[3]{k} \Longleftrightarrow \sqrt[3]n\leq \left(\frac{n}{n-1}\sqrt[3]{\frac{n+1}{n}}-1\right)\sum_{k=1}^{n-1}\sqrt[3]{k}\Longleftrightarrow$ $(!) \sum_{k=1}^{n-1}\sqrt[3]{k}\geq \frac{(n-1)\sqrt[3]{n^2}}{n\sqrt[3]{n+1}-(n-1)\sqrt[3]n}$. Let's prove this using induction on $n$. Base $n=2$ is trivial. Assume true for $n=m$. Let's prove for $n=m+1$. $(!)\frac{m\sqrt[3]{(m+1)^2}}{(m+1)\sqrt[3]{m+2}-m\sqrt[3]{m+1}}\leq \sqrt[3]m+\sum_{k=1}^{m-1}\sqrt[3]{k}$. Using the induction hypothesis we have $(!) \sum_{k=1}^{m-1}\sqrt[3]{k}\geq \frac{(m-1)\sqrt[3]{m^2}}{m\sqrt[3]{m+1}-(m-1)\sqrt[3]m}$. Therefore proving $\frac{m\sqrt[3]{(m+1)^2}}{(m+1)\sqrt[3]{m+2}-m\sqrt[3]{m+1}}\leq \frac{(m-1)\sqrt[3]{m^2}}{m\sqrt[3]{m+1}-(m-1)\sqrt[3]m}+\sqrt[3]m$ is sufficient. $\Longleftrightarrow\frac{\sqrt[3]{m+1}}{(m+1)\sqrt[3]{m+2}-m\sqrt[3]{m+1}}\leq \frac{\sqrt[3]{m}}{m\sqrt[3]{m+1}-(m-1)\sqrt[3]m}$$\Longleftrightarrow\sqrt[3]{m^2(m+1)}-m+1\leq$ $\sqrt[3]{(m+1)^2(m+2)}-m\Longleftrightarrow\sqrt[3]{m^2(m+1)}+1\leq\sqrt[3]{(m+1)^2(m+2)}\Longleftrightarrow(m+1)^3+(m+1)^2\geq m^3+m^2+$ $1+3\sqrt[3]{m^3+m^2}(\sqrt[3]{m^3+m^2}+1)\Longleftrightarrow$ $2m+1\geq3(\sqrt[3]{m^3+m^2}-m)(\sqrt[3]{m^3+m^2}+m+1)$. Let $\sqrt[3]{m^3+m^2}=m+x$. Then $(!)\>2m+1\geq3x(2m+1+x)$. $m^3+m^2=(m+x)^3=m^3+3m^2x+3mx^2+x^3\geq m^3+3m^2x+3mx^2\Longrightarrow m\geq3mx+3x^2\Longrightarrow 1\geq3x,\> $ $2m+1\geq6mx+6x^2+1\geq6mx+6x^2+3x=3x(2m+1+x)$, Q.E.D.