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?
 A: 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
$$\sum_{cyc}(a-b)^2(a+b)\geq0.$$ 
A: Hint:  Just expand and you're done. 
(As per Michael's first step) Expanding out the denominator and homogenizing, we WTS
$$ (a+b+c)(ab+bc+ca) \leq 3 abc + 2(a^3+b^3+c^3)$$
Expanding terms and cancelling $abc$, we WTS   
$$ a^2b + b^2 a + c^2 b + b^2 a + c^2 b + a^2 c \leq 2 (a^3 + b^3 + c^3).$$
This should be obvious / well known e.g. 
$a^2b + b^2 a + c^2 b \leq a^3 + b^3 + c^3$ and $b^2 a + c^2 b + a^2 c \leq a^3 + b^3 + c^3$. 

Unsurprisingly, since you applied an identity (as opposed to an inequality), we can follow the same steps of homogenizing using your approach.      
We WTS
$$ 7 (ab+bc+ca)(a+b+c)   \leq 9 abc + 2(a+b+c)^3$$
Expanding, and cancelling common terms, this becomes:
$$ a^2b + b^2 a + c^2 b + b^2 a + c^2 b + a^2 c \leq 2 (a^3 + b^3 + c^3).$$
