Proving : $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4(a+b+c)$ For $a,b,c > 0$ and $ab+bc+ca+2abc=1$, how to prove that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4(a+b+c) \, ?$$

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $u$, which says that it's enough to prove
our inequality for a maximal value of $u$, which happens for equality case of two variables.
Let $b=a$.
Hence, the condition gives $c=\frac{1-a}{2a}$, where $0<a<1$ and after these substitutions
we need to prove that 
$$\frac{1}{a}+\frac{2a}{1-a}\geq4\left(2a+\frac{1-a}{2a}\right)$$ or
$$(2a-1)^2\geq0.$$
Done!
