Inequality with a High Degree Constraint This question-

Suppose that $x, y, z$ are positive real numbers and $x^5 + y^5 + z^5 = 3$. Prove that $$ {x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \ge 3 $$

The inequality has a high degree constraint which can convert a $5$-degree polynomial to a $0$-degree term and makes it difficult. 
On trying C-S to manage-
$$ \left({x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3}\right)\left(x^5 + y^5 + z^5\right) \ge 9 \Rightarrow   
\left(x^2y+y^2z+z^2x\right)^2\geq9 \Rightarrow x^2y+y^2z+z^2x\geq3 $$Still gives a third degree inequality and not a useful fifth degree.
How can I do it and solve the problem?

We'll  prove that $$x^{6.5}+y^{6.5}+y^{6.5}\geq x^6+y^6+z^6,$$ for which it's enough to prove that:
$$\sum_{cyc}\left(10x^{6.5}-10x^6-x^5+1\right)\geq0$$ or
$$5\sum_{cyc}\left(2x^{1.5}-3x+1\right)x^{5}+\sum_{cyc}\left(5x^6-6x^5+1\right)\geq0,$$ which is true by AM-GM.
Now, by C-S and Vasc we obtain:
$$\sum_{cyc}\frac{x^4}{y^3}=\sum_{cyc}\frac{x^{13}}{x^9y^3}\geq\frac{\left(\sum\limits_{cyc}x^{6.5}\right)^2}{\sum\limits_{cyc}x^9y^3}\geq\frac{\left(\sum\limits_{cyc}x^{6}\right)^2}{\sum\limits_{cyc}x^9y^3}\geq3.$$
The following inequality is also true.

Let $x$, $y$ and $z$ be positive numbers such that $x^{34}+y^{34}+z^{34}=3.$ Prove that:
$$\frac{x^4}{y^3}+\frac{y^4}{z^3}+\frac{z^4}{x^3}\geq3$$

But it's not for contest.
