<s>[INST] Inequality $\frac1a + \frac1b + \frac1c \leq \frac{a^8+b^8+c^8}{a^3b^3c^3}$ 
Let $a,b,c$ be positive reals .
Prove that $\displaystyle \frac1a + \frac1b + \frac1c \leq \frac{a^8+b^8+c^8}{a^3b^3c^3}$

I found this one in a book, no hints mentioned, but marked as very hard.
I can't make any progress...
 [/INST] We might make repeated use of the inequality $x^2+y^2+z^2 \ge xy+yz+zx$, for $x,y,z \in \mathbb{R}$ as well.
We have infact: $$\displaystyle a^8+b^8+c^8 \ge \sum\limits_{cyc} a^4b^4 \ge \sum\limits_{cyc} a^2b^4c^2 \ge \sum\limits_{cyc} ab^2c.bc^2a = \sum\limits_{cyc} a^2b^3c^3 = a^3b^3c^3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$$
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