Prove that if $a+b+c = 1$ then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 3+ 2\frac{a^3+b^3+c^3}{abc}$
What I've done so far is to note that
$$ \begin{align} \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{ab+bc+ca}{abc} \end{align} $$
And by AM-GM along with the fact that $a+b+c = 1$ we have that
$$ \begin{align} ab+bc+ca \leq a^2+b^2+c^2 &= (a^2+b^2+c^2)(a+b+c)\\ &= a^3+b^3+c^3+a^2b+b^2c+c^2a+a^2c+b^2a+c^2b \end{align} $$
And once again using AM-GM we have that $a^2b+b^2c+c^2a \leq a^3+b^3+c^3$ so using this we have that:
$$ \begin{align} \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 2\frac{a^3+b^3+c^3}{abc}+\frac{a^2c+b^2a+c^2b}{abc} \end{align} $$
But I don't know how to proceed from here because the only thing I can think of is that by the same AM-GM argument $a^2c+b^2a+c^2b \leq a^3+b^3+c^3$ but this only leads me to:
$$ \begin{align} \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 3\frac{a^3+b^3+c^3}{abc} \end{align} $$
But that's not what I'm asked to prove so any help would be appreciated.
Multiplying throughout by $abc$ and homogenising, we need to equivalently prove:
$$(ab+bc+ca)(a+b+c) \leqslant 3abc+2(a^3+b^3+c^3)$$ $$\iff (a^2b+b^2c+c^2a)+(ab^2+bc^2+ca^2)\leqslant 2(a^3+b^3+c^3)$$
Which is true, as by Rearrangement Inequality, $a^3+b^3+c^3\geqslant a^2b+b^2c+c^2a$ and $a^3+b^3+c^3\geqslant ab^2+bc^2+ca^2$.