<s>[INST] For positive $a$, $b$, $c$ with $abc=1$, show $(a + b + c)^3\left(\frac1{a^3+8}+\frac1{b^3+8}+\frac1{c^3+8}\right)\geq9$ Prove that for every $a, b, c > 0$ such that $abc = 1$, 
$$ {(a + b + c)} ^ 3 \left(\frac{1}{a ^ 3 + 8} + \frac {1}{b ^ 3 + 8} + \frac {1}{c ^ 3 + 8}\right) \geq 9$$
$$(a + b + c) ^ 3 \left(\frac {1}{a ^ 3 + 8} + \frac {1}{b ^ 3 + 8} + \frac{1}{c ^ 3 + 8})\right) \geq (a + b + c) ^ 3\frac {9}{a ^ 3 + b ^ 3 + c ^ 3 + 24}$$ from Bergstrom's Inequality.
So I need to prove $$(a + b + c) ^ 3 \frac {1}{a ^ 3 + b ^ 3 + c ^ 3 + 24} \geq 1$$ which means $$(a + b + c) ^ 3 \geq a ^ 3 + b ^ 3 + c ^ 3 + 24$$
$$(a + b + c) ^ 3 - a ^ 3 - b ^ 3 - c ^ 3 \geq 24$$
$$ 3(a + b)(b + c)(c + a) \geq 24$$
$$ (a + b)(b + c)(c + a) \geq 8$$
 But        $$a + b \geq 2\sqrt ab$$
$$b + c \geq 2\sqrt bc$$
$$c + a \geq 2\sqrt ca$$
Multiplying these last 3 inequalities, we get $$(a + b)(b + c)(c + a) \geq 8 \sqrt {a^2b^2c^2} = 8 \sqrt 1 = 8$$
 [/INST] By AM-GM and C-S:
$$(a+b+c)^3\sum_{cyc}\frac{1}{a^3+8}=(a+b+c)^3\sum_{cyc}\frac{1}{a^3+8abc}=$$
$$=\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)\sum_{cyc}\frac{1}{a^3+8abc}=$$
$$=\sum_{cyc}(a^3+(a+b)(a+c)(b+c))\sum_{cyc}\frac{1}{a^3+8abc}\geq$$
$$\geq\sum_{cyc}(a^3+8abc)\sum_{cyc}\frac{1}{a^3+8abc}\geq9.$$
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