Set $ a^2 = \frac{yz}{x},\, b^2 = \frac{zx}{y},\, c^2 = \frac{xy}{z}\implies a,b,c>0$ the inequality can be rewritten as \begin{align*} a^2+b^2+c^2 &> \sqrt[3]{a^3b^3+b^3c^3+c^3a^3} \\ \left(a^2+b^2+c^2\right)^3 &> 8\sum_{cyc} a^3b^3 \\ \sum_{cyc} a^6 + 3\sum_{sym} a^4b^2 + 6a^2b^2c^2 &> 8\sum_{cyc} a^3b^3 \end{align*} Schur's inequality implies $ \sum_{cyc} a^6 + 3a^2b^2c^2 \ge \sum_{sym} a^4b^2$ and $ 4\sum_{sym} a^4b^2 \ge 8\sum_{cyc} a^3b^3 = 4\sum_{sym} a^3b^3$ is a consequence of Muirhead (or AM-GM). Using $ 3a^2b^2c^2>0$, we have proven the inequality.

Fang-jh wrote: Let $ x,y,z$ be positive real numbers, show that $ \frac {yz}{x} + \frac {zx}{y} + \frac {xy}{z} > 2\sqrt [3]{x^3 + y^3 + z^3}.$ bertram wrote: Set $ a^2 = \frac {yz}{x},\, b^2 = \frac {zx}{y},\, c^2 = \frac {xy}{z}\implies$ the inequality can be rewritten as $ \left(a^2 + b^2 + c^2\right)^3 \& > 8\left(b^3c^3 + c^3a^3 + a^3b^3\right).$ $ \left(a^2 + b^2 + c^2\right)^3 - 8\left(b^3c^3 + c^3a^3 + a^3b^3\right)\equiv F(a,b,c) = F(a,a + s,a + t)$ $ = 3a^6 + 6(s + t)a^5 + 15(s^2 + t^2)a^4 + 4(s + t)(7s^2 - 10st + 7t^2)a^3$ $ + 3(7s^4 - 10s^2t^2 + 7t^4)a^2 + 6(s - t)^2(s + t)^3a + (s - t)^2(s^4 + 2s^3t + 6s^2t^2 + 2st^3 + t^4) > 0$, which is clearly true for $ 0 < a = \min\{a,b,c\}$. By the way, exist real polynomials $ p_1,p_2,p_3,p_4$ in $ a,b,c$ such that $ F(a,b,c) = p_1^2 + p_2^2 + p_3^2 + p_4^2$, see: http://www.mathlinks.ro/Forum/viewtopic.php?t=202683

Let $a,b,c$ are positive real numbers. Prove that \[\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge \sqrt[3]{8(a^3+b^3+c^3)+3abc}.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2895265

Set $ a=\sqrt{\frac{y^3z^3}{x^3}}, b=\sqrt{\frac{x^3z^3}{y^3}},c=\sqrt{\frac{x^3y^3}{z^3}} $.The inequality can be rewritten as $ \sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}> 2\sqrt[3]{ab+bc+ca} $ Using the holder inequality $ LHS^3(a^2+b^2+c^2)\geq (a+b+c)^4 $.Therefore, it suffices to show that $ \frac{(a+b+c)^4}{a^2+b^2+c^2}\geq 8(ab+bc+ca) $ Using the AM-GM inequality $ 8(ab+bc+ca)(a^2+b^2+c^2)=4(2ab+2bc+2ca)(a^2+b^2+c^2) \leq (a+b+c)^4 $ Q.E.D

Proof by Nguyen Hai Phong Apply $$(x+y+z)^3+9xyz \ge 4(x+y+z)(xy+yz+zx)$$with $x, y, z>0$ we have: $$(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c})^3+9.\frac{bc}{a}.\frac{ca}{b}.\frac{ab}{c}$$ $$\ge 4(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}).(a^2+b^2+c^2)$$ $$=4 \sum a^3(\frac{b}{c}+\frac{c}{b})+12abc \ge 8(a^3+b^3+c^3)+12abc$$by AM-GM so $$(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c})^3 \ge 8(a^3+b^3+c^3)+3abc$$ or $$\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c} \ge \sqrt[3]{8(a^3+b^3+c^3)+3abc}>\sqrt[3]{8(a^3+b^3+c^3)}=2\sqrt[3]{a^3+b^3+c^3}$$ Done.