<s>[INST] How to prove this inequality? $(a+b+c=1)$ Show that if $a,b,c$ are positive reals and $a+b+c=1$, then the following must hold: 
$$\frac{2(a^3+b^3+c^3)}{abc}+3 \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
What I have tried is using $abc \leq \frac{1}{27}$ $(a+b+c \geq 3\sqrt[3]{abc}) $ and multiplying everything by $abc$, but I don'think that's a good idea because $abc$ can be positive and negative.
I have also tried substituting $a^3+b^3+c^3 \geq 3abc$, but that isn't strong enough.
Any help/hints please?? 
 [/INST] if $a,b,c>0$ then we get 
$$\frac{2(a^3+b^3+c^3)}{abc}+3\geq \frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}$$
this is equivalent to
$$\frac{2(a^3+b^3+c^3)}{abc}\geq \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}$$
and this
$$2(a^3+b^3+c^3)\geq bc(b+c)+ac(a+c)+ab(a+b)$$ 
and now note that
$$a^3+b^3=(a+b)(a^2-ab+b^2)\geq ab(a+b)$$
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