Question: If $a,b,c>0$ and $a+b+c=1$, then prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}$.
Solution: Observe that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}\\\iff \frac{ab+bc+ca}{abc}\le 3+2\cdot\frac{a^3+b^3+c^3}{abc}\\\iff ab+bc+ca\le 3abc+2(a^3+b^3+c^3)\hspace{0.5 cm}(\because abc>0)\\\iff(ab+bc+ca)(a+b+c)-3abc\le 2(a^3+b^3+c^3)\hspace{0.5 cm}(\because a+b+c=1)\\\iff a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\le 2(a^3+b^3+c^3).$$
Thus it is sufficient to prove that $$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\le 2(a^3+b^3+c^3).$$
Let us assume WLOG that, $a\le b\le c\implies a^2\le b^2\le c^2.$ Thus by rearrangement inequality, we have $$a^2b+b^2c+c^2a\le a^2a+b^2b+c^2c=a^3+b^3+c^3.\tag 1$$
Again, by rearrangement inequality, we have $$a^2c+b^2a+c^2b\le a^2a+b^2b+c^2c=a^3+b^3+c^3. \tag 2$$
Thus adding $(1)$ and $(2)$, we have $$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\le 2(a^3+b^3+c^3),$$ which is our required inequality. Hence, we are done.
Is there a better solution than this one?
Also, by using the cyclic sum we can use SOS and we can write the solution in one line: $$3+\tfrac{2(a^3+b^3+c^3)}{abc}-\sum_\text{cyc}\tfrac{1}{a} = \tfrac{\sum\limits_\text{cyc}(2a^3+abc-a^2b-a^2c-abc)}{abc} = \tfrac{\sum\limits_\text{cyc}(a^3-a^2b-ab^2+b^3)}{abc} = \tfrac{\sum\limits_\text{cyc}(a+b)(a-b)^2}{abc}\geq0.$$