<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] Hint: Use AM-GM inequality we have
$$3a^8+3b^8+2c^8\ge 8a^3b^3c^2$$
proof: since
$$3a^8+3b^8+2c^8=a^8+a^8+a^8+b^8+b^8+b^8+c^8+c^8\ge 8(a^{8+8+8}b^{8+8+8}c^{8+8})^{1/8}=8a^3b^3c^2$$
similar 
$$3a^8+2b^8+3c^8\ge 8a^3b^2c^3$$
$$2a^8+3b^3+3c^3\ge 8a^2b^3c^3$$
so
$$8(a^8+b^8+c^8)\ge 8a^2b^2c^2(ab+bc+ac)$$
so
$$\dfrac{a^8+b^8+c^8}{a^3b^3c^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$
</s>