I came across this question on Facebook:
Given $a$, $b$, $c$ positive such that $3\geq abc(a+b+c)$, prove that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq a+b+c$$
With reference from other posts, I tried to manipulate the 9 in Cauchy Schwartz inequality: $$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+b+c\right)\geq 9$$ by squaring the constraint and it gets more complicated. Any hints will be much appreciated. Thank you.
