<s>[INST] Inequality question. Let $a,b,c>0$ with $a+b+c=1$. Show that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3 + 2\cdot\frac{\left(a^3 + b^3 + c^3\right)}{abc}$$
Ohhhkk. So first off,
\begin{align} a^3 + b^3+ c^3 & =a^3 + b^3+ c^3- 3abc +3abc\\
& =\ (a+b+c)(a^2+b^2+c^2-(ab+bc+ca))+3abc\\
& = \ (1-3(ab+bc+ca)) + 3abc \\
\end{align}
Using this the inequality becomes,
  $$7 \cdot \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 9+ \frac{2}{abc}$$
How do i proceed from here? Was this the right approach? Is there a better one?
 [/INST] I think the following is a smooth enough.
We need to prove that:
$$(a+b+c)(ab+ac+bc)\leq3abc+2(a^3+b^3+c^2)$$ or
$$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)\geq0,$$ which is obvious.
Your second inequality it's indeed just the first:
$$7\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\leq9+\frac{2}{abc}$$ it's
$$7(a+b+c)(ab+ac+bc)\leq9abc+2(a+b+c)^3$$ or
$$\sum_{cyc}(3abc+2a^3+6a^2b+6a^2c+4abc-7a^2b-7a^2c-7abc)\geq0$$ or
$$\sum_{cyc}(2a^3-a^2b-a^2c)\geq0$$ or
$$\sum_{cyc}(a^3-a^2b-ab^2+b^3)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)\geq0.$$ 
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