<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] Firstly, multiply the right-hand side by $a+b+c$, so that both sides are now degree-zero.  Then you want to prove 
$$2a^3+2b^3+2c^3\geq a^2b+ab^2+a^2c+ac^2+b^2c+bc^2$$
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